- M208_6Pure mathematics
Number systems
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978-1-4730-0615-7 (.epub)IntroductionIn this course we look at some different systems of numbers, and the rules for combining numbers in these systems. For each system we consider the question of which elements have additive and/or multiplicative inverses in the system. We look at solving certain equations in the system, such as linear, quadratic and other polynomial equations.In Section 1 we start by revising the notation used for the **rational numbers** and the **real numbers**, and we list their arithmetical properties. You will meet other properties of these numbers in the analysis units, as the study of real functions depends on properties of the real numbers. We note that some quadratic equations with rational coefficients, such as *x*^{2} = 2, have solutions which are real but not rational.In Section 2 we introduce the set of **complex numbers**. This system of numbers enables us to solve *all* polynomial equations, including those with no real solutions, such as *x*^{2} + 1 = 0. Just as real numbers correspond to points on the real line, so complex numbers correspond to points in a plane, known as the **complex plane**.In Section 3 we look further at some properties of the **integers**, and introduce **modular arithmetic**. This will be useful in the group theory units, as some sets of numbers with the operation of modular addition or modular multiplication form **groups**.In Section 4 we introduce the concept of a **relation** between elements of a set. This is a more general idea than that of a function, and leads us to a mathematical structure known as an **equivalence relation**. An equivalence relation on a set classifies elements of the set, separating them into disjoint subsets called **equivalence classes**.This OpenLearn course is an adapted extract from the Open Unviersity course M208: *Pure Mathematics*After studying this course, you should be able to:understand the arithmetical properties of the rational and real numbersunderstand the definition of a **complex number**perform arithmetical operations with complex numbersexplain the terms modular addition and modular multiplicationexplain the meanings of a relation defined on a set, an equivalent relation and a partition of a set.1 Real numbers1.1 Rational numbersIn OpenLearn course M208_5 Mathematical language you met the sets = {1, 2, 3, …}, the natural numbers; = {…, −2, −1, 0, 1, 2, …}, the integers; = {*p*/*q* : *p* , *q* }, the rational numbers. is the set of numbers that can be written as fractions, such as
and –1.7 =
.Notice that each set in this list is a subset of the succeeding one; that is,
To represent the rational numbers geometrically, we use a number line. We begin by drawing a line and marking on it points corresponding to the integers 0 and 1. If the distance between 0 and 1 is taken as a unit of length, then the rational numbers can be arranged on the line in a natural order. For example, the rational 4/3 is placed at the point which is one third of the distance from 1 to 2.We sometimes use ‘rational’ as shorthand for ‘rational number’ and ‘real’ for ‘real number’.
Each rational number also has a decimal representation, which is either a terminating (that is, finite) decimal, such as 1.54, or a recurring decimal such as 2.134 734 734 7 …, in which the digits repeat in a regular pattern from some position onwards. The decimal representation of any rational number *p*/*q* can be obtained by using long division to divide *q* into *p*.One of the surprising mathematical discoveries made by the ancient Greeks was that the system of rational numbers is not adequate to describe all the lengths that occur in geometry. For example, consider the diagonal of a square of side 1. What is its length? If the length is *x*, then, by Pythagoras' Theorem, *x* must satisfy the equation *x*^{2} = 2. However, there is no rational number that satisfies this equation.
Theorem 1There is no rational number *x* such that *x*^{2} = 2.**Proof** Suppose that such a rational number *x* exists. Then we can write *x* = *p*/*q*, where *p*, *q* . By cancelling, if necessary, we may assume that the greatest common factor of *p* and *q* is 1.This is a proof by contradiction.
The equation *x*^{2} = 2 now becomes so The square of an odd number is odd, so *p* cannot be odd. Hence *p* is even, so we can write *p* = 2*r*, say. Equation (1.1) now becomes
so
We haveReasoning as before, we find that *q* is also even.Since *p* and *q* are both even, they have a common factor of 2, which contradicts our earlier statement that the greatest common factor of *p* and *q* is 1.Since we have obtained a contradiction, our original assumption must be false; therefore no such rational number *x* exists.Exercise 1By imitating the proof above, show that there is no rational number *x* such that *x*^{3} = 2.SolutionSuppose that there exists a rational number *x* such that *x*^{3} = 2. We can write *x* = *p*/*q*, where *p* and *q* are positive integers whose greatest common factor is 1. Then the equation *x*^{3} = 2 becomesNow the cube of an odd number is odd because
so *p* must be even. If we write *p* = 2*r*, then our equation becomes so we have Hence *q* is also even, so 2 is a common factor of *p* and *q*. This contradiction shows that such a number *x* cannot exist.Since we expect equations such as *x*^{2} = 2 and *x*^{3} = 2 to have solutions, we must introduce new numbers that are not rationals. We denote the positive solutions of these two equations by , and
, respectively; thus ()^{2} = 2 and
. Of course, we must introduce many other new numbers, such as , , and so on. Indeed, it can be shown that if *m* and *n* are natural numbers, and the equation *x*^{m} = *n* has no integer solution, then the positive solution of this equation, written as
, cannot be rational.The case *m* = 2 and *n* = 3 is treated in Exercise 5.There are many other mathematical quantities which cannot be described exactly by rational numbers. For example, the number , which denotes the area of a disc of radius 1 (or half the length of the perimeter of such a disc), cannot be described by a rational number, and neither can the number *e*.
Johann Heinrich Lambert proved in 1761 that is irrational. He was a colleague of Euler and Lagrange at the Berlin Academy of Sciences, and is best known for his work on .
All these numbers are known as **irrational** numbers.1.2 Real numbersThe rational and irrational numbers together make up the **real** numbers. The set of real numbers is denoted by . Like rationals, irrational numbers can be represented by decimals, but unlike the decimals for rational numbers, those for irrationals are neither finite nor recurring. All such infinite non-recurring decimals represent real numbers. Each real number can be represented as a point on the number line considered in Section 1.1, which is often known as the **real line**, and each point on this line represents a real number.We shall need to use the usual arithmetical operations on real numbers, and we now list the properties which we assume these operations satisfy.In properties A3 and M3 the numbers −*a* and *a*^{−1} are known as the *additive inverse* (or *negative*) of *a* and the *multiplicative inverse* (or *reciprocal*) of *a*, respectively.The rational numbers also satisfy all the above properties; that is, if is replaced by throughout in the box above, then the properties are still true. We shall show later that the same is true for the complex numbers and for some sets of numbers in modular arithmetic. However, if we restrict ourselves to the integers , then one of these properties is no longer true.
A set with all these properties is known as a *field*.
Exercise 2(a) Show by means of a counter-example that does not have property M3.(b) Which integers have a multiplicative inverse in ?Solution(a) There is no integer 2^{−1} such that 2 × 2^{−1} = 2^{−1} × 2 = 1, for example.(b) The numbers 1 and − 1 both have a multiplicative inverse in . In each case the number is its own inverse; such a number is sometimes described as being *self-inverse*.In this course we shall be concerned with *polynomial equations*.DefinitionA **polynomial equation** in *x* of degree *n* is an equation of the form *p*(*x*) = 0, where *p*(*x*) is a polynomial of degree *n*.Polynomial equations (and polynomials) of degrees 1, 2 and 3 are called *linear*, *quadratic* and *cubic*, respectively.Recall that the formula for the solutions of the quadratic equation *ax*^{2} + *bx* + *c* = 0 isThis equation is known as the *quadratic formula*.Exercise 3Solve the following quadratic equations, stating how many solutions each equation has in .(a) *x*^{2} − 7*x* + 12 = 0(b) *x*^{2} + 6*x* + 9 = 0(c) 2*x*^{2} + 5*x* − 3 = 0(d) 2*x*^{2} − 2*x* − 1 = 0(e) *x*^{2} − 2*x* + 5 = 0Solution(a) This equation has two solutions in .(b) This equation has one solution in .(c) This equation has two solutions in .(d) This equation has two solutions in .(e) Since 4 − 20 = −16, which is negative, this equation has no solutions in .We notice from the results of Exercise 3 that some quadratic equations have two real solutions, some have only one and some have none. In either of the first two cases, the solutions may be rational or irrational. Although you may be accustomed to equations with integer coefficients such as those in Exercise 3, these results still apply if some or all of the coefficients are irrational; that is, if the coefficients are any real numbers. Since a quadratic equation with coefficients in can have no solution in , working with the set of real numbers does not enable us to find solutions to all quadratic equations.We end this section by showing that although a quadratic equation may have no real solutions, this is not true for a cubic equation.Consider the cubic polynomial For *x* ≠ 0, the polynomial can be written as Since
,
and
as , the sign of the polynomial will be positive when *x* is large and positive, and negative when *x* is large and negative. Hence the value of the polynomial must change from negative to positive at least once as *x* increases, so it would appear that it must be zero at one value of *x* (at least). A similar argument can be used if *a* < 0. Hence the equation *ax*^{3} + *bx*^{2} + *cx* + *d* = 0, where *a* ≠ 0, has at least one real solution. This result can be generalised to show that any polynomial equation of degree *n*, where *n* is odd, has at least one real solution.
We are assuming here that the graph of a cubic polynomial is an unbroken curve – an assumption we made in OpenLearn course M208_4 Real functions and graphs. This assumption will be justified in the analysis units.
1.3 Further exercisesExercise 4Solve the following linear equations.(a) 5*x* + 8 = −2(b) *x* + 2 = −5(c) 2*x* −1 = 5(d) 5*x* + 4 = 3State in each case whether the solution belongs to , , and/or .Solution(a) *x* = −2, which belongs to (and hence to and ).(b)
, which belongs to .(c) *x* = 3, which belongs to (and hence to , and ).(d)
, which belongs to (and hence to ).*Remark* Each linear equation with coefficients in has a solution in (because the equation *ax* + *b* = 0 where *a*, *b* and *a* ≠ 0 has solution *x* = − *b*/*a*, which involves only additive inverses and division of rational numbers). Similarly each linear equation with coefficients in has a solution in . The same is not true of , as part (d) illustrates.Exercise 5Show that there is no rational number *x* such that *x*^{2} = 3.SolutionSuppose there is a rational number *x* such that *x*^{2} = 3. Then we can write *x* = *p*/*q*, where *p* and *q* are positive integers whose greatest common factor is 1.Then the equation *x*^{2} = 3 becomes Now *p* is either divisible by 3 or has remainder 1 or 2 on division by 3; that is, *p* = 3*k* or 3*k* + 1 or 3*k* + 2 for some integer *k*.But if *p* = 3*k* + 1, then *p*^{2} = 9*k*^{2} + 6*k* + 1 is not divisible by 3, and if *p* = 3*k* + 2, then is not divisible by 3. So, since 3*q*^{2} is divisible by 3, we conclude that *p* = 3*k*.Hence (3*k*)^{2} = 3*q*^{2}, so *q*^{2} = 3*k*^{2}.But then the same argument applies to *q* to show that *q* must also be divisible by 3.Hence 3 is a common factor of *p* and *q*.This is a contradiction, so we conclude that the assumption must have been false. Hence there is no rational number *x* such that *x*^{2} = 3.2 Complex numbers2.1 What is a complex number?We will now discuss complex numbers and their properties. We will show how they can be represented as points in the plane and state the Fundamental Theorem of Algebra: that any polynomial equation with complex coefficients has a solution which is a complex number. We will also define the function exp of a complex variable.Earlier we mentioned several sets of numbers, including , , and . In each case, the numbers correspond to points on the real line. We now introduce numbers of a new kind, the so-called *complex numbers*, which correspond to points in the plane.Complex numbers arise naturally as solutions of quadratic equations. For example, you showed in Exercise 3(e) that the equation *x*^{2} − 2*x* + 5 = 0 has no real solutions because there is no real number whose square is −16.But suppose now that we ‘extend’ the set of real numbers by introducing a new number, denoted by *i*, which is defined to have the property that *i*^{2} = −1. Suppose further that *i* combines with itself, and with real numbers, according to the usual rules of arithmetic. In particular, assume that we can multiply *i* by any real number *b* to obtain the product *ib*, and that we can add on any real number *a* to obtain the sum *a* + *ib*. Such sums are known as *complex numbers*, and they are the numbers we need to solve the quadratic equation above.
We have *ib* = *bi*, and we can also write *a* + *ib* as *a* + *bi*.
DefinitionsA **complex number** is an expression of the form *x* + *iy*, where *x* and *y* are real numbers and *i*^{2} = −1. The set of all complex numbers is denoted by .A complex number *z* = *x* + *iy* has **real part** *x* and **imaginary part** *y*; we write Two complex numbers are equal when their real parts *and* their imaginary parts are equal.
For example, if *z* = 2 − 3*i*, then Re *z* = 2 and Im *z* = −3.
**Remarks**Any given real number *x* can be written in the form *x* + *i*0, and any complex number of the form *x* + *i*0 is usually written simply as *x*. In this sense, is a subset of . The zero complex number 0 + *i*0 is written as 0.We usually write a general complex number as *x* + *iy*, and a particular complex number as, for example, 2 + 3*i*, rather than 2 + *i*3. We write 2 − 3*i* rather than 2 + (−3)*i*.Note that Re *z* and Im *z* are both real numbers.A complex number of the form 0 + *iy* (where *y* ≠ 0) is sometimes called an *imaginary number*.Since *i*^{2} = −1, we can regard the number *i* as a square root of −1. If we assume that the usual rules of arithmetic apply, then we can solve quadratic equations using the quadratic formula and the fact that
. For example, consider the equation This is the equation from Exercise 3(e), rewritten using *z* as the variable name. We often use the letter *z* for a complex variable (a variable that represents a complex number).For the above equation, the quadratic formula gives We can check that these two complex numbers satisfy the equation we were trying to solve. We use the usual rules of arithmetic, and substitute −1 for *i*^{2} wherever it appears.For example, if *z* = 1 + 2*i*, then
4*i*^{2} = 4(−1) = −4, since *i*^{2} = −1.
The solution *z* = 1 − 2*i* can be checked similarly.The use of the number *i* enables us to solve any quadratic equation in a similar way. We shall see later in the section that the use of *i* ensures that all polynomial equations have solutions, even those whose coefficients are themselves complex. This, in turn, means that any polynomial can be factorised into a product of linear factors; for example,Exercise 6Solve the following equations, giving all solutions in .(a) *z*^{2} − 4*z* + 7 = 0(b) *z*^{2} − *iz* + 2 = 0(c) *z*^{3} − 3*z*^{2} + 4*z* − 2 = 0 (Hint: *z* = 1 is one solution.)(d) *z*^{4} − 16 = 0Solution(a) The equation *z*^{2} − 4*z* + 7 = 0 has solutions that is, *z* = 2 + *i*and *z* = 2 − *i**i*.(b) The equation *z*^{2}− *iz* + 2 = 0 has solutions that is, *z* = 2*i* and *z* = − *i*.(c) We can factorise the equation as Hence *z* = 1 or
,so the solutions are *z* = 1, *z* = 1 + *i* and *z* = 1 − *i*.(d) *z*^{4}− 16 = 0 can be factorised as giving *z*^{2} = 4 or *z*^{2} = −4.Hence *z* = 2 or *z* = −2 or *z* =2*i* or *z* = −2*i*.2.2 The complex planeJust as there is a one-one correspondence between the real numbers and the points on the real line, so there is a one-one correspondence between the complex numbers and the points in the plane. This correspondence is given by Thus we can represent points in the plane by complex numbers and, conversely, we can represent complex numbers by points in the plane. When we represent complex numbers by points in the plane, we refer to the plane as the **complex plane**, and we often refer to the complex numbers as *points* in the complex plane. A diagram showing complex numbers represented as points in the plane in this way is sometimes called an **Argand diagram**.The French mathematician Jean-Robert Argand's publication of the idea in 1806 was the first to be generally recognised.
Real numbers are represented in the complex plane by points on the *x*-axis; this axis is called the *real axis*. Similarly, numbers of the form *iy* are represented by points on the *y*-axis; this axis is called the *imaginary axis*.Exercise 7Draw a diagram showing each of the following points in the complex plane:
Solution
2.3 Complex arithmeticArithmetical operations on complex numbers are carried out as for real numbers, except that we replace *i*^{2} by −1 wherever it occurs.Example 1Let *z*_{1} = 1 + 2*i* and *z*_{2} = 3 − 4*i*. Determine the following complex numbers.(a) *z*_{1} + *z*_{2}(b) *z*_{1} − *z*_{2}(c) *z*_{1}*z*_{2}(d)
SolutionUsing the usual rules of arithmetic, with the additional property that *i*^{2} = −1, we obtain the following.(a) *z*_{1} + *z*_{2} = (1 + 2*i*) + (3 − 4*i*) = (1 + 3) + (2 − 4)*i* = 4 − 2*i*(b) z_{1} − z_{2} = (1 + 2*i*) − (3 − 4*i*) = (1 − 3) + (2 + 4)*i* = − 2 + 6*i*(c) z_{1}z_{2} = (1 + 2*i*)(3 − 4*i*) = 3 + 6*i *− 4*i* − 8*i*^{2} = 3 + 2*i *+ 8 = 11+ 2*i*(d)
Exercise 8Determine the following complex numbers.(a) (3 − 5*i*) + (2 + 4*i*)(b) (2 − 3*i*)(−3 + 2*i*)(c) (5 + 3*i*)^{2}(d) (1 + *i*)(7 + 2*i*)(4 − *i*)Solution(a) (3 − 5*i*) + (2 + 4*i*) = 5 − *i*(b) (2 − 3*i*)(−3 + 2*i*) = −6 + 9*i* + 4*i* − 6*i*^{2} = 13*i*(c) (d) We find so Example 1 illustrates how we add, subtract and multiply two given complex numbers. We can apply the same methods to two general complex numbers *z*_{1} = *x*_{1} + *iy*_{1} and *z*_{2} = *x*_{2} + *iy*_{2}, and obtain the following formal definitions of addition, subtraction and multiplication in .DefinitionsLet *z*_{1} = *x*_{1} + *iy*_{1} and *z*_{2} = *x*_{2} + *iy*_{2} be any complex numbers. Then the following operations can be applied.**addition** *z*_{1} + *z*_{2} = (*x*_{1} + *x*_{2}) + *i*(*y*_{1} + *y*_{2})**subtraction** *z*_{1} − *z*_{2} = (*x*_{1} − *x*_{2}) + *i*(*y*_{1} − *y*_{2})**multiplication** *z*_{1}*z*_{2} = (*x*_{1}*x*_{2} − *y*_{1}*y*_{2}) + *i*(*x*_{1}*y*_{2} + *y*_{1}*x*_{2})With these definitions, most of the usual rules of algebra still hold, as do many of the familiar algebraic identities. For example,andThere is no need to remember or look up these formulas. For calculations, the methods of Example 1 may be used.An obvious omission from this list of definitions is *division*. We return to division after discussing the *complex conjugate* and *modulus* of a complex number.2.4 Complex conjugateMany manipulations involving complex numbers, such as division, can be simplified by using the idea of a *complex conjugate*, which we now introduce.DefinitionThe **complex conjugate**
of the complex number *z* = *x* + *iy* is the complex number *x* − *iy*.
For example, if *z* = 1 − 2*i*, then
. In geometric terms,
is the image of *z* under reflection in the real axis.Exercise 9Let *z*_{1} = −2 + 3*i* and *z*_{2} = 3 − *i*.Write down
and
, and draw a diagram showing *z*_{1}, *z*_{2},
and
in the complex plane.Solution
The following properties of complex conjugates are particularly useful.Properties of complex conjugatesLet *z*_{1}, *z*_{2} and *z* be any complex numbers. Then:
;
;
;
.In order to prove property 1, we consider two arbitrary complex numbers. Let *z*_{1} = *x*_{1} + *iy*_{1} and *z*_{2} = *x*_{2} + *iy*_{2}. Then Exercise 10Use a similar approach to prove properties 2, 3 and 4.Solution*Property 2*Let *z*_{1} = *x*_{1} + *iy*_{1} and *z*_{2} = *x*_{2} + *iy*_{2}. Then so Also,Therefore *Property 3*Let *z* = *x* + *iy*. Then *Property 4*Let *z* = *x* + *iy*. Then 2.5 Modulus of a complex numberWe also need the idea of the *modulus* of a complex number. Recall that the modulus of a real number *x* is defined by
For example, |7| = 7 and |−6| = 6.
In other words, |*x*| is the distance from the point *x* on the real line to the origin. We extend this definition to complex numbers as follows.DefinitionThe **modulus** |*z*| of a complex number *z* is the distance from the point *z* in the complex plane to the origin.Thus the modulus of the complex number *z* = *x* + *iy* is
For example, if *z* = 3 − 4*i*, then
.Exercise 11Determine the modulus of each of the following complex numbers.(a) 5 + 12*i*(b) 1 + *i*(c) −5Solution(a)
(b)
(c)
The modulus of a complex number has many properties similar to those of the modulus of a real number.Properties of modulus|*z*| ≥ 0 for any *z* , with equality only when *z* = 0.|*z*_{1}*z*_{2}| = |*z*_{1}||*z*_{2}| for any *z*_{1}, *z*_{2} .Property 1 is clear from the definition of |*z*|. Property 2 can be proved in a similar way to property 2 of complex conjugates given in Exercise 10.The following useful result shows the link between modulus and distance in the complex plane.Distance FormulaThe distance between the points *z*_{1} and *z*_{2} in the complex plane is |*z*_{1} − *z*_{2}|.This is obtained by applying Pythagoras' Theorem to the triangle in the diagram below.Exercise 12For each of the following pairs *z*_{1}, *z*_{2} of complex numbers, draw a diagram showing *z*_{1} and *z*_{2} in the complex plane, and evaluate |*z*_{1} − *z*_{2}|.(a) *z*_{1} = 3 + *i*, *z*_{2} = 1 + 2*i*.(b) *z*_{1} = 1, *z*_{2} = *i*.(c) *z*_{1} = −5 − 3*i*, *z*_{2} = 2 − 7*i*.Solution(a)
Here so (b)
Here so
(c)
Here so The following properties describe the relationship between the modulus and the complex conjugate of a complex number.Conjugate–modulus properties
for all *z* .
for all *z* .To prove these properties, we let *z* = *x* + *iy*. Then , so
and 2.6 Division of complex numbersThe second of the conjugate–modulus properties enables us to find reciprocals of complex numbers and to divide one complex number by another, as shown in the next example. As for real numbers, we cannot find a reciprocal of zero, nor divide any complex number by zero.Example 2(a) Find the reciprocal of 2 − 5*i*.(b) Find the quotient
Solution(a) We want to find the complex number which represents
We multiply the numerator and denominator by 2 + 5*i*, the complex conjugate of 2 − 5*i*, to give
(b) We multiply the numerator and denominator by 1 −2*i*, the complex conjugate of 1 + 2*i*, to give
The method used in Example 2, of multiplying the numerator and denominator by the complex conjugate of the denominator, enables us to find the reciprocal of any non-zero complex number *z*, and the quotient *z*_{1}/*z*_{2} of any two complex numbers *z*_{1} and *z*_{2}, where *z*_{2} ≠ 0. We can obtain general formulas as follows.For the reciprocal, we have If *z* = *x* + *iy*, so
and |*z*|^{2}= *x*^{2} + *y*^{2}, we obtain For the quotient *z*_{1}/*z*_{2}, we have If *z*_{1} = *x*_{1} + *iy*_{1} and *z*_{2} = *x*_{2} + *iy*_{2}, this can be rewritten as These formulas may be used in theoretical work, but for calculations of reciprocals and quotients it is simplest to use the method of Example 2.Exercise 13Find the reciprocal of each of the following complex numbers.(a) 3 − *i*(b) −1 + 2*i*SolutionIn each case we multiply both the numerator and the denominator by the complex conjugate of the denominator.(a)
(b)
Exercise 14Evaluate each of the following quotients.(a)
(b)
SolutionIn each case we multiply the numerator and denominator by the complex conjugate of the denominator.(a)
(b)
2.7 Arithmetical properties of complex numbersThe set of complex numbers satisfies all the properties previously given for arithmetic in . We state (but do not prove) these properties here.In particular, 0 = 0 + 0*i* plays the same role in as the real number 0 does in , and 1 = 1 + 0*i* plays the same role as 1. These numbers are called *identities* for addition and multiplication respectively.We also have that the *additive inverse* (or negative) of *z* = *x* + *yi* is −*z* = −*x* − *yi*, and the *multiplicative inverse* (or reciprocal) of *z* = *x* + *yi* is There is one important difference between the set of real numbers and the set of complex numbers, however; namely that, unlike the real numbers, the complex numbers are not ordered.For any two real numbers *a* and *b*, exactly one of the three properties is true. But this is not the case for the complex numbers; we cannot say, for example, that
Inequalities involving complex numbers make sense only if they are inequalities between real quantities, such as the moduli (moduli is the plural of modulus) of the complex numbers. For example, inequalities such as are valid.2.8 Polar formYou have seen that the complex number *x* + *iy* corresponds to the point (*x*, *y*) in the complex plane. This correspondence enables us to give an alternative description of complex numbers, using so-called *polar form*. This form is particularly useful when we discuss properties related to multiplication and division of complex numbers.
Polar form is obtained by noting that the point in the complex plane associated with the non-zero complex number *z* = *x* + *iy* is uniquely determined by the modulus
,together with the angle *θ* (measured in an anticlockwise direction in radians) between the positive direction of the *x*-axis and the line from the origin to the point, as shown in the graph above. We have so the complex number *z* can be expressed as This description of *z* in terms of *r* and *θ* is not unique because the angle *θ* is determined only up to multiples of 2; that is, the angles *θ* ± 2, *θ *± 4, *θ* ± 6, …, also determine the same complex number. However, if we restrict the angle *θ *to lie in the interval (−, ], then the description *is* unique. The origin is an exception, however: at the origin the value of *r* is 0, and *θ* is not defined.
Some texts restrict *θ* to lie in the interval [0, 2).
DefinitionsA non-zero complex number *z* = *x* + *iy* is in **polar form** if it is expressed as where *r* = |*z*| and *θ* is any angle (measured in radians anticlockwise) between the positive direction of the *x*-axis and the line joining *z* to the origin.
(Some texts use *r* cis *θ* or 〈*r*, *θ* 〉 as shorthand for *r*(cos *θ* + *i* sin *θ*).)
Such an angle *θ* is called an **argument** of the complex number *z*, and is denoted by arg *z*. The **principal argument** of *z* is the value of arg *z* that lies in the interval (−, ], and is denoted by Arg *z*.Here *principal argument* is a shortened form of the more conventional ‘principal *value of the* argument’.**Remark** Sometimes we refer to *z* = *x* + *iy* as the *Cartesian form* of *z*, to distinguish it from polar form.We now discuss how to convert a complex number from polar form to Cartesian form, and vice versa.When carrying out such conversions, it is useful to remember the values in the table below, as these will help you in some special cases. You may find it easier to remember the triangles, from which you can work out most of the values in the table.The following formulas are also helpful. For any *θ * :
You will need only the first two equations in each row; the other equations are given for completeness.
You may be able to remember these formulas by roughly sketching graphs of the sine and cosine functions, and using their symmetry. For example, we can sketch the sine function as follows.To convert a complex number from polar form to Cartesian form is straightforward: we use the equations to find *x* and *y* given *r* and *θ*.
This is illustrated in the following example.Example 3Express each of the following complex numbers in Cartesian form.(a) 3(cos(/3) + *i* sin(/3))(b) cos(−/6) + *i* sin(−/6)Solution(a) Here *r* = 3 and *θ* = /3. Thus the required form is *x* + *iy*, where and The Cartesian form is therefore
.(b) Here *r* = 1 and *θ* = −/6. Thus the required form is *x* + *iy*, where and The Cartesian form is therefore
.Exercise 15Express each of the following complex numbers in Cartesian form.(a) 2(cos(/2) + *i* sin(/2))(b) 4(cos(−2/3) + *i* sin(−2/3))Solution(a) The required form is *x* + *iy*, where and The Cartesian form is therefore 2*i*.(b) The required form is *x* + *iy*, where and The Cartesian form is therefore − 2 − 2*i*.To convert a non-zero complex number *z* from Cartesian form *x* + *iy* to polar form *r*(cos *θ *+ *i* sin *θ*), we first find the modulus *r* using the formula If *z* is either real or imaginary, then it lies on one of the axes and has principal argument 0, /2, or −/2, as shown in diagram (a) below. Otherwise, to find the principal argument *θ*, we need to solve the equations We can do this by first finding the first-quadrant angle *φ *that satisfies the related equation The relationship of *φ* to the principal argument* θ *depends on the quadrant in which *z* lies, as indicated in diagram (b) below.
*Note:* *φ* is the angle at the origin in the right-angled triangle formed by drawing the perpendicular from *z* to the real axis, as illustrated above, in the case where *z* lies in the second quadrant. We haveand the relationship between *θ* and *φ* can be seen to be as given in diagram (b).The quadrant in which *z* lies can be found from the values of *x* and *y*, and *θ* can then be deduced by using the appropriate equation from diagram (b). (You may find it helpful to sketch *z* on an Argand diagram.)This method for finding *θ *is used in the next example, which illustrates how the Cartesian form of a complex number is converted into polar form.Example 4Express each of the following complex numbers in polar form, using the principal argument.(a) 2 + 2*i*(b)
Solution(a) Let *z* = *x* + *iy* = 2 + 2*i*, so *x* = 2 and *y* = 2.Then *z* = *r*(cos *θ *+ *i* sin *θ*), where To find* θ*, we calculate So *φ* = /4, and *z* lies in the first quadrant so *θ *= *φ* = /4.
The polar form of 2 + 2*i* in terms of the principal argument is therefore (b) Let
, so
and
.Then *z* = *r*(cos *θ* + *i* sin *θ*), where To find *θ*, we calculate So *φ* = /3, and *z* lies in the third quadrant so *θ *= −( − *φ*) = −2/3.The polar form of
in terms of the principal argument is therefore
Exercise 16Draw a diagram showing each of the following complex numbers in the complex plane, and express them in polar form, using the principal argument.(a) −1 + *i*(b) 1 − *i*(c) −5Solution(a) Let *z* = *x* + *iy* = −1 + *i*, so *x* = −1 and *y* = 1. Then *z* = *r*(cos *θ* + *i* sin *θ*), where Also So *φ* = /4, and *z* lies in the second quadrant, so *θ* = − *φ* = 3/4.Thus the polar form of −1 + *i* in terms of the principal argument is (b) Let *z* = *x* + *iy* = 1 − *i*, so *x* = 1 and *y* = − . Then *z* = *r*(cos *θ* + *i* sin *θ)*, where Also So *φ *= /3, and *z* lies in the fourth quadrant, so *θ* = − *φ *= − /3.Thus the polar form of 1 − √3*i* in terms of the principal argument is (c) Let *z* = *x* + *iy* = −5, so *x* = −5 and *y* = 0. Then *z* = *r*(cos *θ* + *i* sin *θ*), where Also *z* lies on the negative half of the real axis, so *θ* = .Thus the polar form of −5 in terms of the principal argument is Polar form gives us a simple way to multiply complex numbers. Letthen (Here we use the addition formulas for the trigonometric functions.)That is, to multiply two complex numbers in polar form, we multiply their moduli and add their arguments. For example,*Note:* In the rest of this section we usually write /4 as
, and similarly for other fractions of . You may use either form, or the form
, as you wish.We can also use formula (2.1) for the product of two complex numbers in polar form to establish a similar formula for the *quotient* of two complex numbers. Specifically, we show that if with *z*_{2} ≠ 0, then *z*_{1}/*z*_{2} is the complex number *z*_{2} ≠ 0 *r*_{2} ≠ 0To see this, notice that since *r*_{1} = *rr*_{2} and *θ*_{1} = *θ* + *θ*_{2} it follows from the above discussion that *z*_{1} = *zz*_{2}. Hence *z*_{1}/*z*_{2} = *z*, as required. We can write the formula as That is, to divide a complex number *z*_{1} by another complex number *z*_{2}, we divide the modulus of *z*_{1} by the modulus of *z*_{2}, and subtract the argument of *z*_{2} from the argument of *z*_{1}.For example,In particular, if *z* = *r*(cos *θ* + *i* sin *θ*) with *r* ≠ 0, then the reciprocal of *z* is The above methods for multiplying and dividing complex numbers in polar form are summarised below.Strategy 1To multiply two complex numbers given in polar form, multiply their moduli and add their arguments.To divide a complex number *z*_{1} by a non-zero complex number *z*_{2} when both are given in polar form, divide the modulus of *z*_{1} by the modulus of *z*_{2}, and subtract the argument of *z*_{2} from the argument of *z*_{1}.**Remark**If you require the *principal* argument of the product or quotient, then you may need to add or subtract 2 from the argument calculated.Exercise 17Determine each of the following products and quotients in polar form in terms of the principal argument.(a)
(b)
(c)
(d)
Solution(a) The modulus of the product is An argument is Since this argument lies in (− , ], it follows that it is the principal argument. The required product is therefore (b) The modulus of the product is An argument is The principal argument is therefore The required product is therefore (c) The modulus of the quotient is An argument is The principal argument is therefore The required quotient is therefore (d) The modulus of the quotient is An argument is Since this argument lies in (− , ], it follows that it is the principal argument. The required quotient is therefore 2.9 Roots of polynomialsWe begin by reminding you of what we mean by the word ‘root’. In this course we use this term in two different, but related, senses, as given below.DefinitionIf *p*(*z*) is a polynomial, then the solutions of the polynomial equation *p*(*z*) = 0 are called the **roots** of *p*(*z*).If *a* is a complex number, then the solutions of the equation *z*^{n} = *a* are called the *n*th **roots** of *a*.
The roots of a polynomial are also called its *zeros*.
Thus the *n*th roots of *a* are the roots of the polynomial *z*^{n} − *a*.
In this subsection we explain how to find the *n*th roots of any complex number and we discuss the roots of polynomial equations more generally.The formula obtained in the previous section for finding the product of two complex numbers in polar form can be generalised to a product of several complex numbers.Strategy 2To find the product of the complex numbers *z*_{1}, *z*_{2}, …, *z*_{n} given in polar form, multiply their moduli and add their arguments.Exercise 18Let *z*_{1} = −1 + *i*,
and *z*_{3} = −5.Use the solution to Exercise 16 to express *z*_{1}*z*_{2}*z*_{3} and
in polar form.SolutionFrom the solution to Exercise 16,Hence using the principal argument.Also An important special case of the result in Strategy 2 is obtained when *z*_{1} = *z*_{2} = … = *z*_{n} = *r*(cos *θ* + *i* sin *θ*); that is, when the complex numbers *z*_{1}, *z*_{2}, …, *z*_{n} are all equal. In this case, the result gives *Note:* *n**θ* may not be the *principal* argument of (cos *θ* + *i*sin *θ*)^{n}.
Example 5Find *z*^{4}, where *z* = 1 + *i*.SolutionSo *z*^{4} = −4.The significant new information obtained from equation (2.4) is that This result is true for all *n* and is known as *de Moivre's Theorem*.
Abraham de Moivre was a French mathematician who worked in England from 1685 after the expulsion of the Huguenots from France.
Theorem 2: de Moivre's TheoremIf *z* = cos *θ *+ *i* sin* θ*, then, for any *n* ,We have seen that de Moivre's Theorem is true when *n* is a positive integer. We now show that it is true for other integers.For *n* = 0,For *n* = −*m*, where *m* is a positive integer,
Here we use formula (2.3).
One application of de Moivre's Theorem is in finding the *n*th roots of complex numbers; that is, in solving equations of the form *z*^{n} = *a*, where *a* . Before we illustrate this, we use the theorem to verify some solutions of such an equation.Exercise 19(a) Express the complex number 1 in polar form.(b) Show that each of the three complex numbers with polar forms 1(cos 0 + *i* sin 0), and
satisfies the equation *z*^{3} = 1.(c) Express in Cartesian form the three solutions to the equation *z*^{3} = 1 given in part (b).Solution(a) 1 = 1(cos 0 + *i* sin 0)(b) (c) In Cartesian form,
The solution to Exercise 19 verifies that the three given complex numbers are solutions of the equation *z*^{3} = 1. However, what we really want is a method which will enable us to find solutions of such an equation without knowing them in advance. Fortunately, de Moivre's Theorem enables us to do this.Example 6(a) Express the equation *z*^{3}= −27 in polar form.(b) Find three different solutions of the equation *z*^{3}= −27, by using the fact that adding any multiple of 2 to an argument of −27 gives another argument of −27.Solution(a) In polar form, −27 = 27(cos + *i* sin ). If *z* = *r*(cos *θ* + *i* sin *θ*), then the equation *z*^{3}= −27 can be written as (b) From equation (2.5), *r*^{3} = 27, so *r* = 3.
Also from equation (2.5), cos 3*θ* = cos and sin 3*θ* = sin , so one solution for *θ* is obtained by taking 3*θ* = , so
.However, we could also take 3, 5, 7, … as arguments of −27, so 3*θ* = 3, 3*θ *= 5, 3*θ* = 7, … also give solutions. Hence we can have ....But
, so taking
gives the same complex number as
.Similarly, increasing the value of the argument of −27 by any further multiple of 2 repeats solutions we already have.So the three different solutions of *z*^{3} = −27 are given by
These solutions can also be written in Cartesian form as They are shown in the diagram below.
The solution *z*_{3} could be rewritten in terms of its principal argument as 3(cos(−/3) + *i* sin(−/3)).
Now we show how we can use the method of Example 6 to find the solutions of any complex equation of the form where *a* is a known complex number. We write both *z* and *a* in polar form so that, say,where *r* and *θ* are variables whose values we must find, and *ρ *and *φ* are known real numbers.Then *z*^{n} = *a* gives, using de Moivre's Theorem,
Hence we must have *r*^{n} = *ρ*, so *r* = *ρ*^{1/n}. Also *n**θ* must represent the same angle as *φ*. Now we use the fact that a complex number has many arguments. Since adding on any multiple of 2 to the argument *φ* of *a* gives the same complex number *a*, we can have that is,If *k* = *n* we have *θ* = *φ*/*n* + 2, which is the same angle as *φ*/*n*. So taking *k* = 0, 1, 2, …, *n* − 1 will give the *n* different solutions of the equation *z*^{n} = *a*. We thus arrive at the following conclusion.Roots of a complex numberLet *a* = *ρ(*cos *φ *+ *i* sin *φ*) be a complex number in polar form. Then, for any *n* , the equation *z*^{n} = *a* has *n* solutions, given by for *k* = 0, 1, …, *n* − 1.Exercise 20(a) Express the equation *z*^{6} = 1 in polar form.(b) Use the method described above to find the six solutions of *z*^{6} = 1 in polar form.(c) Sketch the position of each solution in the complex plane.(d) Find the Cartesian form of each solution.Solution(a) Let *z* = *r*(cos *θ* + *i* sin *θ*). Then, since we have (b) Hence *r* = 1^{1/6} = 1 and for *k* = 0, 1, … 5, and the six solutions of *z*^{6} = 1 are given byHence the solutions are (c)
(d) In Exercise 20 you found the solutions of the equation *z*^{6} = 1. These are known as the sixth roots of unity, and in the complex plane they are equally spaced around the circle of radius 1, centre the origin. More generally, the solutions of the equation *z*^{n} = 1 are known as the *n***th roots of unity** and in the complex plane they are equally spaced around the circle of radius 1, centre the origin. For any *n* , one of the *n*th roots of unity is 1.The *n*th roots of any complex number are equally spaced around a circle with centre the origin, but the circle may not have radius 1 and there may not be a root on the real axis.Exercise 21Solve the equation *z*^{4} = −4, expressing your answers in Cartesian form. Mark your solutions on a diagram of the complex plane.SolutionLet *z* = *r*(cos *θ* + *i* sin *θ*). Then, since we have Hence *r* = 4^{1/4} = and
for *k* = 0, 1, 2, 3.So the solutions are
Exercise 22Solve the equation *z*^{3} = 8*i*, expressing your answers in Cartesian form. Mark your solutions on a diagram of the complex plane.SolutionLet *z* = *r*(cos *θ* + *i* sin *θ*).Since
, we haveHence *r* = 8^{1/3} = 2 and
for *k* = 0, 1, 2.So the solutions are
The result in the box above gives the *n* solutions of any equation of the form *z*^{n} = *a*, where *a* is a non-zero complex number. Now the equation *z*^{n} = *a*, which can be written as *z*^{n} − *a* = 0, is an example of a polynomial equation whose coefficients are complex numbers. Other examples of polynomial equations with complex coefficients are and Remarkably, it can be shown that *any* polynomial equation where *a*_{n}, *a*_{n−1}, …, *a*_{0} and *a*_{n} ≠ 0, has at least one solution in . Moreover, it cannot have more than *n* solutions.*Note:* If we count coincident solutions separately so that, for example, in the equationthe solution 1 is counted three times, the solution −4 is counted twice and the solution 5 is counted once, then a polynomial equation of degree *n* has *exactly n* solutions.The complex numbers are, unlike the reals, an ‘algebraically closed’ system of numbers. By this we mean that any polynomial equation with coefficients in has a solution in . The fact that any polynomial equation with complex coefficients necessarily has a complex solution is called the *Fundamental Theorem of Algebra*. We do not prove this theorem in this course.It is often not easy to find such solutions! However, in some cases we can combine the methods of the Polynomial Factorisation Theorem below and the use of de Moivre's Theorem to find solutions of a polynomial equation.You met the Polynomial Factorisation Theorem for real roots of a polynomial in OpenLearn course M208_5 Mathematical language. The theorem is also true in .Theorem 3: Polynomial Factorisation TheoremLet *p*(*z*) be a polynomial of degree *n* with coefficients in and let . Then *p*() = 0 if and only if where *q*(*z*) is a polynomial of degree *n*−1 with coefficients in .In this statement of the theorem, we have used *z* in place of *x*, as this is the label usually used for a complex variable. The proof is otherwise exactly the same as the proof for the theorem in , so we do not include it here.From the Fundamental Theorem of Algebra, mathematical induction and the Polynomial Factorisation Theorem, we can deduce the following important corollary.CorollaryEvery polynomial *p*(*z*) = *a*_{n}*z*^{n} + *a*_{n}_{−1}*z*^{n}^{−1} + ··· + *a*_{1}*z* + *a*_{0}, where *n* ≥ 1, *a*_{i} for each *i* and *a*_{n} ≠ 0, has a factorisation where the complex numbers _{1}, _{2}, …, _{n} are the roots (not necessarily distinct) of *p*(*z*).**Proof**Let *S*(*n*) be the statement of the corollary for general *n* ≥ 1.*Note:* *S*(*n*) is the statement in the box above. We use the notation *S*(*n*) instead of the more usual *P*(*n*) to avoid confusion with the polynomial *p*(*z*).Then *S*(1) is true, since the polynomial (where *a*_{1} ≠ 0) has the root _{1} = −*a*_{0}/*a*_{1} and has a factorisation Now suppose that *S*(*k*) is true, and consider any polynomial of degree *k* + 1:where *a*_{k+1} ≠ 0. By the Fundamental Theorem of Algebra, the equation *p*(*z*) = 0 has at least one solution, say _{k+1}, in . Then, by the Polynomial Factorisation Theorem,where *q*(*z*) is a polynomial of degree *k*. Now the coefficient of *z*^{k} in *q*(*z*) must be *a*_{k+1}. Thus, by *S*(*k*), this polynomial has a factorisation Thus Therefore *S*(*k*) true *S*(*k* + 1) true, so by mathematical induction *S*(*n*) is true for every positive integer *n*. You may have noticed that it follows from the quadratic formula (see Exercise 3) that, for a quadratic equation with *real* coefficients, the roots are either both real or occur as a complex conjugate pair.This holds for polynomial equations in general; if *p*(*z*) is a polynomial in *z* with real coefficients, then whenever is a root of *p*, so is
. (We omit the proof of this result.) Moreover, the factors *z* − and
can be combined to give a real quadratic factor of *p*(*z*), namely which has real coefficients, since
and
.Example 7(a) Show that *z* = *i* is a root of the polynomial (b) Hence find all the roots of *p*(*z*).Solution(a)
so *i* is a root of *p*(*z*).(b) Since *p* has real coefficients, *z* = −*i* is also a root of *p*(*z*), so (*z* −*i*)(*z* + *i*) = *z*^{2} + 1 is a factor of *p*(*z*). By equating coefficients, we obtain So the remaining two roots of *p*(*z*) are given by the solutions of the equation *z*^{2} − 3*z* + 1 = 0.Using the quadratic formula, we have Hence the four roots of *p*(*z*) are *i*, −*i*,
and
.Exercise 23Find, in the form *a*_{n}*z*^{n} + … + *a*_{1}*z* + *a*_{0}, a polynomial whose roots are 1, −2, 3*i* and −3*i*.SolutionA suitable polynomial is that is,or Earlier in this subsection we used de Moivre's Theorem to find roots of complex numbers. Another use of de Moivre's Theorem is to find further trigonometric identities similar to those given in OpenLearn course M208_4 Real functions and graphs.Exercise 24Using de Moivre's Theorem, we havefor all *θ* .(a) Expand the left-hand side of the above expression using the Binomial Theorem. Then express your answer in the form *x* + *iy*, where *x* and *y* are expressions involving cos *θ* and sin *θ*.(b) By equating real and imaginary parts, use your answer to part (a) to obtain formulas for cos 3*θ* and sin 3*θ* in terms of cos *θ* and sin *θ*.(c) Use your answer to part (b) and the identity cos^{2} *θ* + sin^{2} *θ *= 1 to obtain a formula for cos 3*θ* in terms of cos *θ* and a formula for sin 3*θ* in terms of sin *θ*.Solution(a) (b) By part (a),so and (c) Since sin^{2} *θ* = 1 − cos^{2} *θ *and cos^{2} *θ* = 1 − sin^{2} *θ*, we have so and so The method of the solution to Exercise 24 generalises to produce formulas for cos *n**θ* and sin *n**θ* for all *n* .Exercise 25(a) Use de Moivre's Theorem to obtain formulas for cos 5*θ* and sin 5*θ* in terms of cos *θ* and sin *θ*.(b) Use your answer to part (a) and the identity cos^{2} *θ* + sin^{2} *θ* = 1 to find a formula for cos 5*θ* in terms of cos *θ* and a formula for sin 5*θ* in terms of sin *θ*.Solution(a) Equating real and imaginary parts gives and (b) Since sin^{2} *θ* = 1 − cos^{2} *θ* and sin^{4} *θ* = (sin^{2} *θ*)^{2} =(1 − cos^{2} *θ*)^{2}, on substituting into the formula for cos 5*θ *found in part (b) we obtain so Similarly,so 2.10 The complex exponential functionConsider the real exponential function *f* (*x*) = *e*^{x} (that is, *f* (*x*) = exp *x*). We now extend the definition of this function to define a function *f*(*z*) = *e*^{z} whose domain and codomain are .We expect complex powers of *e* to satisfy the familiar properties of real powers of *e*. For example, we expect that If this is to be achieved, then the definition of *e*^{z} has to be as follows.DefinitionIf *z* = *x* + *iy*, then *e*^{z} = *e*^{x}*e*^{iy} = *e*^{x}(cos *y* + *i* sin *y*).Example 8Show thatfor all complex numbers *z*_{1} and *z*_{2}.SolutionWe use Strategy 1.Suppose that *z*_{1} = *x*_{1} + *iy*_{1} and *z*_{2} = *x*_{2} + *iy*_{2}. Then We assume that if *x*_{1}, *x*_{2} are *real* numbers, thenExercise 26(a) Using the above definition for *e*^{z} and de Moivre's Theorem, show that (b) Show that *e*^{z1}/ *e*^{z2} = *e*^{z1−z2}, for all *z*_{1}, *z*_{2} .Solution(a) Let *z* = *x* + *iy*; then (by de Moivre's Theorem with *n* = − 1)(b) So the rules for multiplication and division of complex powers of *e* are exactly the same as those for real powers. Furthermore, when the exponent *z* is real, that is when *z* = *x* + 0*i*, where *x* , the definitions of a real and a complex power of *e* coincide, since On the other hand, if *z* = 0 + *iy*, where *y* , then the definition gives This is known as **Euler's formula**. Putting *y* = , we obtain or This is a remarkable relationship between five important numbers: 0, 1, *i*, and *e*.The formula *e*^{iy} = cos *y* + *i* sin *y* gives us an alternative form for the expression of a complex number in polar form. If then we can write cos *θ* + *i* sin *θ *as *e*^{iθ}, so Some texts refer to *re*^{iθ} as *polar* form.A complex number expressed in this way is said to be in **exponential form**. Using this notation, de Moivre's Theorem becomes the simple result We can use the complex exponential function to find some further useful trigonometric identities. At the end of Section 2.9 we showed how de Moivre's Theorem can be used to express the sine or cosine of a multiple of* θ* in terms of powers of sin *θ *and cos *θ*. Here we do the opposite, expressing a power of sin *θ* or cos *θ* as a combination of sines or cosines of multiples of *θ*.First, we deduce two useful equations. We know that, for all *θ* ,Also, since cos(−*θ*) + *i* sin(−*θ*)=cos *θ *− *i* sin *θ*, we have Adding equations 2.6 and 2.7 gives and subtracting them gives Equations 2.8 and 2.9 enable us to express cos *θ* and sin *θ* as combinations of complex exponentials, and it is these equations we use to obtain the identities we are looking for.Example 9(a) Show that
.(b) Expand the expression (*e*^{iθ} + *e*^{−iθ})^{4} using the Binomial Theorem, and hence show that Solution(a) 2cos *θ* = *e*^{iθ} + *e*^{−iθ}, so (2 cos *θ*)^{4} = (*e*^{iθ} + *e*^{iθ})^{4}.Hence
.(b) Hence Using Equation 2.8 first with 4*θ *in place of *θ* and then with 2*θ* in place of *θ*, we have as required.Exercise 27(a) Show that
.(b) Expand (*e*^{iθ} − *e*^{−iθ})^{5} using the Binomial Theorem, and hence show that
.Solution(a) 2*i* sin *θ* = *e*^{iθ} − *e*^{− iθ}, so (2*i* sin *θ*)^{5} =(*e*^{i θ} − *e*^{−i θ})^{5}.Hence (b) Hence 2.11 Further exercisesExercise 28Let *z*_{1} = 2 + 3*i* and *z*_{2} = 1 − 4*i*. Find *z*_{1} + *z*_{2}, *z*_{1} − *z*_{2}, *z*_{1}*z*_{2},
,
, *z*_{1}/*z*_{2} and 1/*z*_{1}.Solution
Exercise 29Draw a diagram showing each of the following complex numbers in the complex plane, and express them in polar form, using principal arguments.(a) − *i*(b) −5*i*(c) −2 − 2*i*Solution
**(a)** Let *z* = *x* + *iy* = − *i*, so *x* = and *y* = − 1. Then *z* = *r*(cos *θ* + *i* sin *θ*), where Also
. So
, and *z* lies in the fourth quadrant, so
. Hence the polar form of − *i* in terms of the principal argument is (b) Let *z* = *x* + *iy* = −5*i*, so *x* = 0 and *y* = −5. Then *z* = *r*(cos *θ* + *i* sin *θ*), where Also *z* lies on the negative half of the imaginary axis, so
.Hence the polar form of −5*i* in terms of the principal argument is (c) Let *z* = *x* + *iy* = −2 − 2*i*, so *x* = −2 and *y* = −2. Then *z* = *r*(cos *θ* + *i* sin *θ*), where Also
. So
, and *z* lies in the third quadrant, so
.Hence the polar form of −2 − 2*i* in terms of the principal argument is Exercise 30Express each of the following complex numbers in Cartesian form.(a)
(b)
(c)
Solution(a) The required form is *x* + *iy*, where and so the Cartesian form is 2 + 2*i*.(b) The required form is *x* + *iy*, where so the Cartesian form is 3*i*.(c) The required form is *x* + *iy*, where and so the Cartesian form is
Exercise 31Let *z*_{1} = − *i*, *z*_{2} = −5*i* and *z*_{3} = −2 −2*i*. Use the solution to Exercise 29 to determine the following complex numbers in polar form in terms of the principal argument.(a) *z*_{1}*z*_{2}*z*_{3}(b)
SolutionFrom the solution to Exercise 29, we have **(a)** Hence using the principal argument.**(b)** Exercise 32Solve the equation *z*^{5} = −32, leaving your answers in polar form.SolutionLet *z* = *r*(cos *θ* + *i* sin *θ*).Then, since −32 = 32(cos + *i* sin ), we have Hence *r* = 2 and
for any integer *k*, and the five solutions of *z*^{5} = −32 are given by for *k* = 0, 1, 2, 3, 4.Hence the solutions are Exercise 33Solve the equation *z*^{3} + *z*^{2} − *z* + 15 = 0, given that one solution is an integer.SolutionThe integer solution must be a factor of the constant term 15,so it must be one of ± 1, ± 3, ± 5, ± 15.Testing these, we find *z* = −3 is a root, since Hence *z* + 3 is a factor, and we find that The solutions of *z*^{2} −2*z* + 5 = 0 are given by Hence the solutions of *z*^{3} + *z*^{2} − *z* + 15 = 0 are *z* = −3, *z* =1 + 2*i* and *z* = 1 − 2*i*.Exercise 34Determine a polynomial of degree 4 whose roots are 3, −2, 2 − *i* and 2 + *i*.SolutionA suitable polynomial is that is,or Exercise 35Use de Moivre's Theorem to obtain formulas for cos 6*θ* and sin 6*θ* in terms of cos *θ* and sin *θ*.SolutionHence and Exercise 36Use the definition of *e*^{z} to express the following complex numbers in Cartesian form.(a) (b) (c) Solution(a) (b) (c) = *e*^{−1}(cos + *i*sin) = −*e*^{−1}.3 Modular arithmetic3.1 DivisionIn this section, instead of enlarging the number system , we do arithmetic with finite sets of integers, namely the sets of possible remainders when we divide by particular positive integers. This type of arithmetic is important in number theory (the study of the integers) and in cryptography. It is used frequently in group theory.If we divide one positive integer by another we obtain a **quotient** and a **remainder**. For example, 29 divided by 4 gives quotient 7 and remainder 1 because 29 = 7 × 4 + 1. If we divide any positive integer by 4, the remainder will be one of the numbers 0, 1, 2, 3.This idea can be extended to the division of a negative integer by a positive integer. For example, −19 divided by 4 gives quotient −5 and remainder 1 because −19 = (−5) × 4 + 1. If we divide any negative integer by 4, the remainder is again one of the numbers 0, 1, 2, 3.This result can be generalised to the following theorem.Theorem 4 Division AlgorithmLet *a* and *n* be integers, with *n* > 0. Then there are unique integers *q* and *r* such that
Strictly speaking, this theorem is not an algorithm, but ‘division algorithm’ is the traditional name for it.We say that dividing *a* by the **divisor** *n* gives **quotient** *q* and **remainder** *r*.We do not give a formal proof of Theorem 4, but it can be illustrated as follows. We mark integer multiples of *n* along the number line as shown in the diagram below, and then observe in which of the resulting intervals of length *n* the integer *a* lies. Suppose that *a* lies in the interval [*qn*, (*q* +1)*n*) so that *qn* ≤ *a* < (*q* + 1)*n*.
Then, if we let *r* = *a* − *qn*, we have *a* = *qn* + *r* and 0 ≤ *r* < *n*, which is the required result.Exercise 37For each of the following integers *a* and *n*, find the quotient and remainder on division of *a* by *n*.(a) *a* = 65, *n* =7(b) *a* = −256, *n* =13Solution(a) 65 = 9 ×7 + 2, so the quotient is 9 and the remainder is 2.(b) −256 = −20 × 13 + 4, so the quotient is −20 and the remainder is 4.Exercise 38(a) What are the possible remainders on division of an integer by 7?(b) Find two positive and two negative integers all of which have remainder 3 on division by 7.Solution(a) The possible remainders are 0, 1, 2, 3, 4, 5 and 6.(b) There are many possible answers here; for example, 3, 10, −4 and −11.3.2 CongruenceThe Division Algorithm tells us that, when we divide any integer by a positive integer *n*, the set of possible remainders is {0, 1, 2, …, *n* − 1}. Integers which differ by a multiple of *n* have the same remainder on division by *n* and are, in this sense, ‘the same’ as each other. We now introduce some notation and terminology for this idea of ‘sameness’, which is known as *congruence*.DefinitionsLet *n* be a positive integer. Two integers *a* and *b* are **congruent modulo ***n* if *a* − *b* is a multiple of *n*; that is, if *a* and *b* have the same remainder on division by *n*.In symbols we write (We say ‘*a* is congruent to *b* modulo *n*’.)Such a statement is called a **congruence**, and *n* is called the **modulus** of the congruence.*Note:* This is a different meaning for the word ‘modulus’ from others you may have met. It is important to interpret technical terms according to their context.Example 10Which of the following congruences are true, and which are false?(a) 27 ≡ 5 (mod 11)(b) 14 ≡ −6 (mod 3)(c) 343 ≡ 207 (mod 68)(d) 1 ≡ −1 (mod 2)Solution(a) 27 − 5 = 22, which is a multiple of 11, so this congruence is true.Alternatively, we could note that 27 = 2 ×11 + 5 and 5 = 0 ×11 + 5, so 27 and 5 both have remainder 5 on division by 11.(b) 14 − (−6) = 20, which is not a multiple of 3, so this congruence is false.Alternatively, 14 = 4 × 3 + 2 and −6 = (−2) × 3 + 0, so 14 has remainder 2 on division by 3, but −6 has remainder 0.(c) 343 − 207 = 136 = 2 × 68, so this congruence is true.Alternatively, 343 = 5 × 68 + 3 and 207 = 3 ×68 + 3, so 343 and 207 both have remainder 3 on division by 68.(d) 1 − (−1) = 2, so this congruence is true.Both 1 and −1 have remainder 1 on division by 2.It is often simplest to check a congruence *a* ≡ *b* (mod *n*) by considering the difference *a* − *b*.Exercise 39Find the remainder on division by 17 of each of the numbers 25, 53, −15, 3 and 127, and state any congruences modulo 17 that exist between these numbers.SolutionWe have so the remainders are 8, 2, 2, 3 and 8, respectively. So 25 ≡ 127 (mod 17) and 53 ≡ −15 (mod 17).We shall need to use some properties of congruences in the following sections, and we state these properties here.Theorem 5 Properties of congruencesLet *n* and *k* be positive integers, and let *a*, *b*, *c*, *d* be integers. Then(a) *a* ≡ *a* (mod *n*);(b) if *a* ≡ *b* (mod *n*), then *b* ≡ *a* (mod *n*);(c) if *a* ≡ *b* (mod *n*) and *b* ≡ *c* (mod *n*), then *a* ≡ *c* (mod *n*);(d) if *a* ≡ *b* (mod *n*) and *c* ≡ *d* (mod *n*), then *a* + *c* ≡ *b* + *d* (mod *n*);(e) if *a* ≡ *b* (mod *n*) and *c* ≡ *d* (mod *n*), then *ac* ≡ *bd* (mod *n*);(f) if *a* ≡ *b* (mod *n*), then *a*^{k} ≡ *b*^{k} (mod *n*).Although this may seem a long list, these properties are quite simple.**Proof**We prove properties (a)–(d) here and ask you to prove properties (e) and (f) in Exercise 40.(a) *a* − *a* = 0 = 0 × *n*, so *a* ≡ *a* (mod *n*).(b) Suppose that *a* ≡ *b* (mod *n*). Then *a* − *b* = *kn* for some integer *k*. Hence *b* − *a* = (−*k*)*n*, so *b* ≡ *a* (mod *n*).(c) Suppose that *a* ≡ *b* (mod *n*) and *b* ≡ *c* (mod *n*). Then *a* − *b* = *kn* and *b* − *c* = *ln* for some integers *k* and *l*. Hence so *a* ≡ *c* (mod *n*).(d) If *a* ≡ *b* (mod *n*) and *c* ≡ *d* (mod *n*), then *a* − *b* = *kn* and *c* − *d* = *ln* for some integers *k* and *l*. Hence *a* = *b* + *kn* and *c* = *d* + *ln*, so Hence (*a* + *c*) − (*b* + *d*) = (*k* + *l*)*n*, so *a* + *c* ≡ *b* + *d* (mod *n*).Exercise 40(a) Use a similar method to that used in part (d) of the above proof to prove property (e) of Theorem 5.(b) Use property (e) and mathematical induction to prove property (f) of Theorem 5.Solution(a) Suppose that *a* ≡ *b* (mod *n*) and *c* ≡ *d* (mod *n*). Then *a* = *b* + *kn* and *c* = *d* + *ln* for some integers *k* and *l*.Hence so *ac* − *bd* = (*bl* + *kd* + *kln*)*n*, so *ac* ≡ *bd* (mod *n*).(b) Suppose that *a* ≡ *b* (mod *n*) and let *P*(*k*) be the statement *a*^{k} ≡ *b*^{k} (mod *n*).Then *P*(1) is true, since we know that *a* ≡ *b* (mod *n*). Suppose that *r* ≥ 1 and *P*(*r*) is true, that is, *a*^{r} ≡ *b*^{r} (mod *n*). Then, by property (e), since *a*^{r} ≡ *b*^{r} and *a* ≡ *b* (mod *n*), we have *a*^{r} × *a* ≡ *b*^{r} × *b* (mod *n*), that is, Hence *P*(*r*) true *P*(*r* + 1) true, so by mathematical induction *P*(*k*) is true for all positive integers *k*.The properties in Theorem 5 are particularly useful when finding the remainder of a large integer on division by another integer.Example 11(a) Find the remainders of both 2375 and 5421 on division by 22.(b) Find the remainder of 2375 × 5421 on division by 22.(c) Find the remainder of (2375)^{15} on division by 22.(2375)^{15} is too large to fit into the memory of most computers, so there is a real advantage in this method.Solution(a) Using property (c), we can add or subtract any convenient multiples of 22, and make a list of congruences. We have (here we have subtracted 2200, then 110, then 66; then added 22)and (here we have subtracted 4400, then 880, then 110, then 22)so 2375 has remainder 21 on division by 22, and 5421 has remainder 9 on division by 22.(b) Using property (e), we obtain so 2375 × 5421 has remainder 13 on division by 22.(c) Using property (f), we obtain so (2375)^{15} has remainder 21 on division by 22.3.3 Operations in modular arithmeticThe Division Algorithm tells us that all the possible remainders on division by an integer *n* lie in the set We denote this set by _{n}. For each integer *n* ≥ 2 we have a set _{n}, and it is on these sets that we perform *modular arithmetic*. The modular addition operations +_{n} and modular multiplication operations ×_{n} are defined as follows.DefinitionsFor any integer *n* ≥ 2, For *a* and *b* in _{n}, the operations +_{n} and ×_{n} are defined by:*a* +_{n} *b* is the remainder of *a* + *b* on division by *n*;*a* ×_{n} *b* is the remainder of *a* × *b* on division by *n*.The integer *n* is called the **modulus** for this arithmetic.*Note: **a* +_{n} *b* is read as *a* plus *b* (mod *n*), and may also be written *a* + *b* (mod *n*). Similarly, *a* ×_{n} *b* is read as *a* times *b* (mod *n*) and may also be written as *a* × *b* (mod *n*).For example, _{7} = { 0,1,2,3,4,5,6} and we have You have certainly met some modular arithmetic before, as the operations +_{12} and +_{24} are used in measuring time on 12-hour and 24-hour clocks, respectively.Exercise 41Evaluate the following.(a) 3 +_{5} 7, 4 +_{17} 5, 8 +_{16} 12.(b) 3 ×_{5} 7, 4 ×_{17} 5, 8 ×_{16} 12.Solution(a) 3 +_{5} 7 = 0, 4 +_{17} 5 = 9, 8 +_{16} 12 = 4.(b) 3 ×_{5} 7 = 1, 4 ×_{17} 5 = 3, 8 ×_{16} 12 = 0.In Sections 1 and 2 we listed some properties satisfied by the real and complex numbers. We now investigate whether the sets _{n} satisfy similar properties.We also investigate what equations we can solve in _{n}; for example, can we solve the equations These may look much simpler than the equations that we were trying to solve in , but they pose interesting questions. We shall see that the answers may depend on the modulus that we are using.Before we discuss these questions further, we look at addition and multiplication tables, which provide a convenient way of studying addition and multiplication in _{n}.We consider addition first. Here are the addition tables for _{4} and _{7}.
In order to evaluate 4 +_{7} 2, say, we look in the row labelled 4 and the column labelled 2 in the second table to obtain the answer 6.Exercise 42(a) Use the tables above to solve the following equations.
(i) *x* +_{4} 3 = 2
(ii) *x* +_{7} 5 = 2
(iii) *x* +_{4} 2 = 0
(iv) *x* +_{7} 5 = 0
(b) What patterns do you notice in the tables?Solution(a)
(i) 3 +_{4} 3 = 2, so *x* = 3.
(ii) 4 +_{7} 5 = 2, so *x* = 4.
(iii) 2 +_{4} 2 = 0, so *x* = 2.
(iv) 2 +_{7} 5 = 0, so *x* = 2.
(b) You may have noticed that:
each element appears exactly once in each row and exactly once in each column;
there is a pattern of diagonal stripes running down from right to left.
Exercise 43(a) Construct the addition table for _{6}.(b) Solve the equations *x* +_{6} 1 = 5 and *x* +_{6} 5 = 1.Solution(a)
(b) *x* +_{6} 1 = 5 has solution *x* = 4. *x* +_{6} 5 = 1 has solution *x* = 2.For every integer *n* ≥ 2, the additive properties of _{n} are the same as the additive properties of , as follows.Property A1 follows from the Division Algorithm and the definition of _{n}. The other properties can be deduced from the corresponding properties for integers.Exercise 44By using the corresponding property for integers, prove property A5.SolutionBy definition, *a* +_{n} *b* and *b* +_{n} *a* are the remainders of *a* + *b* and *b* + *a*, respectively, on division by *n*. But *a* + *b* = *b* + *a*, so *a* +_{n} *b* = *b*+_{n} *a*.If *a*, *b* _{n} and *a* +_{n} *b* = 0, then we say that *b* is the **additive inverse** of *a* in _{n}. For example, 4 and 5 belong to _{9} and 4 +_{9} 5 = 0, so 5 is the additive inverse of 4 in _{9}. Property A3 states that every element of _{n} has an additive inverse in _{n}.Additive inverses are sometimes written in the form −_{n}*a*; that is, if *a* +_{n} *b* = 0, then we write *b* = −_{n}*a*. For example, 5 = −_{9}4.Exercise 45(a) Use the addition table for _{7} (which appear above Exercise 42) to complete the following table of additive inverses in _{7}.
(b) Complete the following table of additive inverses in _{n}, explaining why your answers are correct.
Solution(a)
(b)
The additive inverse of 0 is always 0, since 0 +_{n} 0 = 0. For any integer *r* > 0 in _{n}, *n*− *r* _{n} and *r* + (*n* − *r*) = *n*, so *r* +_{n} (*n* − *r*) = 0.The existence of additive inverses means that, as well as doing addition modulo *n*, we can also do subtraction. We define *a* −_{n} *b* or, equivalently, *a* − *b* (mod *n*), to be the remainder of *a* − *b* on division by *n*.(With this definition, *a* −_{n} *b* is equal to *a* +_{n} (−_{n}*b*).)For example, to find 2 −_{8} 7, we have Since 3 _{8}, it follows that 3.4 Modular multiplicationIn the last subsection we stated that, for any integer *n* ≥ 2, the set _{n} satisfies the same rules for addition modulo *n* as the real numbers satisfy for ordinary addition. When it comes to multiplication in _{n}, *most* of the familiar rules for multiplication of the real numbers are true. In particular, the following properties hold.The following property also holds.These properties can be proved in a similar way to the additive properties. You will notice that one property is missing from the list of multiplicative properties; namely, the multiplicative inverse property M3.We say that *b* is the **multiplicative inverse** of *a* in _{n} if *a*, *b* _{n} and *a* ×_{n} *b* = *b* ×_{n} *a* = 1. These equations can also be written as *ab* ≡ *ba* ≡ 1 (mod *n*). We denote the multiplicative inverse *b* of *a* by *a*^{−}^{1}, when it exists, and it may be referred to as the multiplicative inverse of *a* modulo *n*. We now investigate the existence of multiplicative inverses.(For example, from the table for _{7} below, 3 ×_{7} 5 = 5 ×_{7} 3 = 1, so 5 is the multiplicative inverse of 3 in _{7}.)Here are the multiplication tables for _{4} and _{7}.
Exercise 46(a) Use the tables above to answer the following.
(i) Which integers in _{4} have multiplicative inverses?
(ii) Find the multiplicative inverse of every integer in _{7} except 0.
(b) Construct a multiplication table for _{10}, and decide which integers in _{10} have multiplicative inverses.Solution(a)
(i) 1 and 3 have multiplicative inverses in _{4}, since 1 ×_{4} 1 = 1 and 3 ×_{4} 3 = 1.
(ii) The multiplicative inverses in _{7} are given by the following table, where *b* is the multiplicative inverse of *a*.
(b)
The integers 1, 3, 7 and 9 have multiplicative inverses in _{10}.From the solution to Exercise 46, it is clear that, unlike and , some systems _{n} contain non-zero elements that do not have a multiplicative inverse. The question of whether an element *a* _{n} has a multiplicative inverse in _{n} is connected with the *common factors* of *a* and *n*.DefinitionsTwo positive integers *a* and *b* have a **common factor** *c*, where *c* is a positive integer, if *a* and *b* are both divisible by *c*.Two positive integers *a* and *b* are said to be **coprime**, or **relatively prime**, if their only common factor is 1.(The *greatest* common factor of *a* and *b* is the largest of their common factors.)Later in this subsection we prove that an element *a* of _{n} has a multiplicative inverse in _{n} if and only if *a* and *n* are coprime. First we look at a method for finding multiplicative inverses where they exist. Although we can find such inverses by trial and error, or by writing out the multiplication table for _{n}, as *n* becomes larger the method illustrated in the following example becomes more efficient. It is known as **Euclid's Algorithm**, and was described in Euclid's *Elements*, which dates from around 300 BC. (Euclid did not express the algorithm in this form, however.)Example 12Find the multiplicative inverse of 10 in _{27}.SolutionWe apply the Division Algorithm repeatedly, starting by dividing the modulus 27 by the integer 10, whose multiplicative inverse we seek:
*Note:* The first part of the method, consisting of repeated application of the Division Algorithm, is the procedure known as Euclid's Algorithm. When it is applied to any two positive integers *a* and *b*, the last but one remainder is the greatest common factor of *a* and *b*. The second part of the method, which involves starting with the last but one equation from the first part, is often described as working backwards through Euclid's Algorithm.At each step we divide the *divisor* in the row above by the *remainder* in the row above, repeating the process until we reach a remainder of 0 (which must occur because the remainders decrease by at least 1 at each step).Now we use Equations 3.1 to find the required multiplicative inverse. Starting with the last but one equation, and working upwards, we have (We rearrange each equation so that only the remainder is on the left-hand side.)We write down the first of Equations 3.2, and use the other two equations to eliminate multiples of 3 and 7 by successive substitutions.(The multiples may be negative; for example, −8 × 10 is a multiple of 10.)After each substitution, 1 is expressed as a multiple of one integer plus a multiple of another integer, where the two integers are a neighbouring pair from the list 27, 10, 7, 3 of left-hand sides of Equations 3.1. The last equation expresses 1 in terms of multiples of the two integers that we started with, 10 and 27. Rearranging this equation gives so Now the integer −8 does not belong to _{27}, but, since −8 ≡ 19 (mod 27), we have and hence Hence 19 is the multiplicative inverse of 10 in _{27} and we can write 10^{−1} = 19 (mod 27).As a check:Exercise 47Use Euclid's Algorithm to find(a) the multiplicative inverse of 7 in _{16};(b) the multiplicative inverse of 8 in _{51}.Solution(a) Starting with the last equation, we have Hence 7 × 7 = 3 × 16 + 1, so 7 ×_{16} 7 = 1 and the multiplicative inverse of 7 in _{16} is 7.(b) Starting with the last equation, we have Hence (−19) × 8 ≡ 1 (mod 51), so Hence 32 ×_{51} 8 = 1 and the multiplicative inverse of 8 in _{51} is 32.We now use Euclid's Algorithm to prove our main result.Theorem 6Let *n* and *a* be positive integers, with *a* in _{n}. Then *a* has a multiplicative inverse in _{n} if and only if *a* and *n* are coprime.**Proof**First we prove the ‘if’ part; that is, *a* has a multiplicative inverse in _{n} if *a* and *n* are coprime.We use Euclid's Algorithm repeatedly and show that, if *a* and *n* are coprime, then the final remainder before we reach 0 must be 1.Let Since the remainders decrease by at least 1 with each step, they must eventually reach 0.The final equation shows that *r*_{m} is a factor of *r*_{m}_{−1}, and thus the penultimate equation shows that *r*_{m} is a factor of *r*_{m}_{−2}, and so on. Continuing in this way, we find that *r*_{m} is a factor of all the remainders, and so of both *a* and *n*. Since *a* and *n* were assumed to be coprime, we deduce that *r*_{m} = 1.Therefore we have, from the penultimate equation, and, by successively substituting for the remainders, we find that there are integers *k* and *d* such that 1 = *kn* + *da*. Hence *da* = −*kn* + 1, so *d* ×_{n} *a* = 1.It is possible that *d* does not belong to _{n}, but in that case *d* ≡ *b* for some *b* _{n}, where *b* ≠ 0, so we also have *b* ×_{n} *a* = 1.Hence *a* has a multiplicative inverse *b* in _{n}; we write *b* = *a*^{−}^{1} (mod *n*).Now we prove the ‘only if ’ part; that is, *a* has a multiplicative inverse in _{n} only if *a* and *n* are coprime.Suppose that *a* has a multiplicative inverse in _{n}; that is, there is a number *b* such that *b* ×_{n} *a* = 1. Then *ba* = *kn* + 1 for some integer *k*, so *ba* − *kn* = 1.If *a* and *n* have a common factor *c*, say, then *c* is a factor of *ba* − *kn* and hence of 1. Therefore *c* can only be 1, so *a* and *n* are coprime.Theorem 6 gives us an important corollary in the case when the modulus *n* is a prime number.CorollaryLet *p* be a prime number. Then every non-zero element in _{p} has a multiplicative inverse in _{p}.**Proof**If *p* is prime, then every non-zero element in _{p} is coprime with *p*, and so has a multiplicative inverse in _{p} by Theorem 6.It follows that if we take multiplication in _{p}, where *p* is prime, then we can add the following property to the list of properties for multiplication in _{n}:M3. If *a* _{p}, and *a* ≠ 0, then *a* has a multiplicative inverse *a*^{−1} _{p}such that *a* ×_{p} *a*^{−1} = *a*^{−1} ×_{p} *a* = 1.But this property does not hold for _{n} if *n* is not prime, as in that case some elements *a* _{n} do not have multiplicative inverses.We now return briefly to the question of whether we can solve equations in modular arithmetic. We begin by considering linear equations, that is, equations of the form where *a*, *c* _{n}. We seek all solutions *x* _{n}.First we consider the case where *a* and *n* are coprime. In this case, by Theorem 6, *a* has a multiplicative inverse *a*^{−}^{1} and we can solve Equation 3.3 by multiplying both sides by this inverse.Example 13Solve the equation 10 ×_{27} *x* = 14.SolutionIn Example 12 we found that the multiplicative inverse of 10 in _{27} is 19. Hence we have so the given equation has the unique solution *x* = 23.In general, by an argument similar to that of Example 13, if *a* and *n* are coprime, then Equation 3.3 has the *unique* solution *x* = *a*^{−1} ×_{n} *c*.(In particular, if *a* and *n* are coprime, then Equation 3.3 has a solution for *every c* _{n}, so *every* element of _{n} appears in the row labelled *a* of the multiplication table for _{n}.)Exercise 48Use the method of Example 13 and the solution to Exercise 47 to solve the following equations.(a) 7 ×_{16} *x* = 3(b) 8 ×_{51} *x* = 19Solution(a) Since the multiplicative inverse of 7 in _{16} is 7 (see the solution to Exercise 47(a)), we have (b) Since the multiplicative inverse of 8 in _{51} is 32 (see the solution to Exercise 47(b)), we have To use the method of Example 13, first we need to find the multiplicative inverse in _{n} of the coefficient *a* of *x*. If we have not already found this inverse (for example, by using Euclid's Algorithm), and the modulus *n* is fairly small, then the quickest way to solve the equation may be just to try different values of *x*. We know that there is a unique solution, so we can stop trying values once we have found a solution. Sometimes a solution can be spotted by using conguences.Example 14Solve the equation 5 ×_{12} *x* =7.SolutionObserve that 7 ≡ −5 (mod 12), so we have The integer −1 is not an element of _{12}, but −1 ≡ 11 (mod 12), so Hence the solution of the given equation is *x* = 11.Now consider Equation 3.3 in the case where *a* and *n* are not coprime: suppose that *a* and *n* have a common factor *d* ≥ 2. Then Equation 3.3 has a solution only if *d* is also a factor of *c*. To see this, notice that Equation 3.3 is equivalent to (Thus if *a* and *n* have a common factor *d* ≥ 2, then all the elements in the row labelled *a* of the multiplication table for _{n} have common factor *d*.)If there exists an integer solution *x* = *b* of this equation, then and, since *d* is a factor of both *a* and *n*, it follows that *d* is a factor of *c*.If *a*, *n* and *c* have a common factor *d* ≥ 2, then Equation 3.3 has more than one solution. In fact, although we do not prove it here, if *d* is the *greatest* common factor of *a*, *n* and *c*, then Equation 3.3 has *d* solutions, given by where *x* = *b* is the smallest solution. That is, we add multiples of *n*/*d* to the smallest solution.*Note:* You can prove that these are solutions by using substitution. For example, *x* = *b* + *n*/*d* is a solution becausewhere the last but one line follows because *x* = *b* is a solution and *a*/*d* _{n}.There is a method for finding the smallest solution which is similar to the method used in Example 13 for the case where *a* and *n* are coprime, but we do not cover it in this course. If *n* is fairly small, then we can find the smallest solution by trying values. Alternatively, it may be possible to spot a solution, but as this may not be the smallest solution, we may need to subtract multiples of *n*/*d* as well as, or instead of, adding them.Example 15Solve each of the following equations.(a) 4 ×_{12} *x* = 6(b) 4 ×_{12} *x* = 8Solution(a) This equation has no solutions, since 4 is a factor of both 4 and 12 but is not a factor of 6.(b) One solution of this equation is *x* = 2. Also *n*/*d* = 12/4 = 3, so the other solutions are *x* = 2 + 3 = 5 and *x* = 2 + 2 × 3 = 8.Exercise 49Find all the solutions of the following equations.(a) In _{12}: 3 ×_{12} *x* = 6, 8 ×_{12} *x* = 7, 5 ×_{12} *x* = 2.(b) In _{16}: 4 ×_{16} *x* = 12, 3 ×_{16} *x* = 13, 8 ×_{16} *x* = 2.Solution(a) One solution of the equation 3 ×_{12} *x* = 6 is *x* = 2. Also *n*/*d* = 12/3 = 4, so the other solutions are *x* = 2 + 4 = 6 and *x* = 2 + 2 × 4 = 10.The equation 8 ×_{12} *x* = 7 has no solutions because 8 and 12 have common factor 4 but 7 does not.Because 5 and 12 are coprime, the equation 5 ×_{12} *x* = 2 has a unique solution. The solution, *x* = 10, can be found in various ways: for example, by calculating *x* = 5^{−1} ×_{12} 2, or by testing possible values for *x*, or by spotting that 2 ≡ −10 (mod 12) and using the fact that the congruence 5 × (−2) ≡ −10 (mod 12) implies 5 × 10 ≡ 2 (mod 12) giving 5 ×_{12} 10 = 2.(b) One solution of the equation 4 ×_{16} *x* =12 is *x* = 3. Also *n*/*d* = 16/4 = 4, so the other solutions are *x* = 3 + 4 = 7, *x* =3 + 8 = 11 and *x* = 3 + 12 = 15.Because 3 and 16 are coprime, the equation 3 ×_{16} *x* = 13 has a unique solution. The solution, *x* = 15, can be found in various ways: for example, by calculating *x* = 3^{−1} ×_{16} 13, or by testing possible values for *x*, or by spotting that 13 ≡ −3 (mod 16) and using the fact that the congruence 3 × (−1) ≡ −3 (mod 16) implies 3 × 15 ≡ 13 (mod 16) giving 3 ×_{16} 15 = 13.The equation 8 ×_{16} *x* = 2 has no solutions because 8 and 16 have common factor 4 but 2 does not.Near the beginning of Section 3.3 we posed the question: can we solve the equation *x* ×_{9} *x* = 2? That is, is there an element of _{9} whose square 2?Example 16Show that there is no element of _{9} whose square is 2.SolutionWe solve this problem by exhaustion, by writing down the squares of all the elements of _{9}.
We can see from this table that there is no element of _{9} whose square is 2. In fact, we can go further and say that the remainder of a square modulo 9 can be only 0, 1, 4 or 7. That is, 0, 1, 4 and 7 are the only elements of _{9} that are squares of other elements.Exercise 50(a) Find all the solutions of the equation *x* ×_{8} *x* = 4.(b) Find all the values of *c* in _{8} for which it is possible to solve the equation *x* ×_{8} *x* = *c*.SolutionWe find all the values of *x* ×_{8} *x*.
(a) The solutions of *x* ×_{8} *x* = 4 are *x* = 2 and *x* = 6.(b) The equation *x* ×_{8} *x* = *c* can be solved for *c* = 0, 1, 4.In general, the solution of quadratic equations in modular arithmetic is more complicated than that of linear equations. You will study this topic further if you take a course in number theory.3.5 Further exercisesExercise 51Evaluate the following sums and products in modular arithmetic.(a) 21 +_{26} 15, 21 ×_{26} 15.(b) 19 +_{33} 14, 19 ×_{33} 14.Solution(a) 21 +_{26} 15 = 10, 21 ×_{26} 15 = 3.(b) 19 +_{33} 14 = 0, 19 ×_{33} 14 = 2.Exercise 52Use Euclid's Algorithm to find:(a) the multiplicative inverse of 8 in _{21};(b) the multiplicative inverse of 19 in _{33}.Solution(a) Hence Hence 8 × 8 = 3 × 21 + 1, so so the multiplicative inverse of 8 in _{21} is 8.(b) Hence Hence so so the multiplicative inverse of 19 in _{33} is 7.Exercise 53Construct the multiplication table for _{11}, and hence find the multiplicative inverse of every non-zero element in _{11}.Solution
Hence we have the following multiplicative inverses in _{11}.
Exercise 54Use the solution to Exercise 52 to solve the following equations.(a) 8 ×_{21} *x* = 13(b) 19 ×_{33} *x* = 15Solution(a) We have 8 ×_{21} *x* = 13. Multiplying by 8, which is the multiplicative inverse of 8 mod 21 (see the solution to Exercise 52(a)), we have so (b) We have 19 ×_{33} *x* = 15. Multiplying by 7, which is the multiplicative inverse of 19 mod 33 (see the solution to Exercise 52(b)), we have so Exercise 55Find all the solutions of the following equations.(a) In _{8}: 3 ×_{8} *x* = 7, 4 ×_{8} *x* = 7, 4 ×_{8} *x* = 4.(b) In _{15}: 3 ×_{15} *x* = 6, 4 ×_{15} *x* = 3, 5 ×_{15} *x* = 2.Solution(a) Because 3 and 8 are coprime, the equation 3 ×_{8} *x* = 7 has a unique solution. The solution, *x* = 5, can be found in various ways: for example, by calculating *x* = 3^{−1} ×_{8} 7, or by testing possible values for *x*, or by spotting that 7 ≡ 15 (mod 8) and using the fact that the congruence 3 × 5 ≡ 15 (mod 8) implies 3 × 5 ≡ 7 (mod 8) giving 3 ×_{8} 5 = 7.The equation 4 ×_{8} *x* = 7 has no solutions because 4 and 8 have common factor 4 but 7 does not.One solution of the equation 4 ×_{8} *x* = 4 is *x* = 1. Also *n*/*d* = 8/4 = 2, so the other solutions are *x* = 1 + 2 = 3, *x* = 1 + 4 = 5 and *x* = 1 + 6 = 7.(b) One solution of the equation 3 ×_{15} *x* = 6 is *x* = 2. Also *n*/*d* = 15/3 = 5, so the other solutions are *x* = 2 + 5 = 7 and *x* = 2 + 10 = 12.Because 4 and 15 are coprime, the equation 4 ×_{15} *x* = 3 has a unique solution. The solution, *x* = 12, can be found in various ways: for example, by calculating *x* = 4^{−1} ×_{15} 3, or by testing possible values for *x*, or by spotting that 3 ≡ −12 (mod 15) and using the fact that the congruence 4 × (−3) ≡ −12 (mod 15) implies 4 × 12 ≡ 3 (mod 15) giving 4 ×_{15} 12 = 3.The equation 5 ×_{15} *x* = 2 has no solutions because 5 and 15 have common factor 5 but 2 does not.Exercise 56(a) Show that the equation *x* ×_{12} *x* = 7 has no solutions.(b) Find all the solutions of *x* ×_{12} *x* = 4.SolutionWe find all the values of *x* ×_{12} *x*.
(a) Hence there is no integer *x* _{12} such that *x* ×_{12} *x* = 7.(b) The solutions of *x* ×_{12} *x* = 4 are *x* = 2, 4, 8, 10.4 Equivalence relations4.1 What is a relation?In this final section we look at a method of classifying the elements of a set by sorting them into subsets. We shall require that the set is sorted into disjoint subsets – so each element of the set belongs to *exactly one* subset. Such a classification is known as a *partition* of a set. In order to achieve a partition, we need to have a method which enables us to decide whether or not one element belongs to the same subset as another. We look first at the general idea of a *relation*, and then at the particular properties needed by a relation in order to partition a set. A relation which satisfies these special properties is known as an *equivalence relation*, and the subsets into which the set is partitioned are called *equivalence classes*.
Equivalence relations occur in all branches of mathematics. For example, in geometry, two possible relations between the set of all triangles in the plane are *is congruent to* and *is similar to*. These are both equivalence relations: the relation *is congruent to* partitions the set of triangles into classes such that all the triangles within each class are congruent to each other, whereas *is similar to* partitions the set into classes of similar triangles. These partitions are different: triangles of the same shape but different sizes are similar to, but not congruent to, each other.Equivalence relations are not confined to sets of mathematical objects. For example, relations between people such as *is the same height as* and *has the same birthday as* are equivalence relations.RelationsWe shall use the symbol (known as tilde or twiddle) to represent a *relation* between two elements of a set.Some texts use *ρ*, rather than , for an arbitrary relation. Certain relations have special symbols; for example,< means *is less than*,= means *is equal to*.DefinitionWe say that is a **relation** on a set *X* if, whenever *x*, *y* *X*, the statement *x* *y* is either true or false.If *x* *y* is true, then *x* is related to *y*.If *x* *y* is false, then *x* is not related to *y* and we write *x* *y*.The statement *x* *y* can be read as ‘*x* is related to *y*’ or ‘*x* twiddles *y*’.Examples1. The condition ‘is equal to’ is a relation on any set of real numbers because, for any *x*, *y* in the set, the statement ‘*x* is equal to *y*’ is either definitely true or definitely false. This relation is usually denoted by the symbol =. For this relation, each real number in the set is related only to itself!2. The condition ‘is less than’ is a relation on any set of real numbers, and we usually denote it by the symbol <. For example, −2 < 1, but 1 −2 and 3 3.3. The condition ‘is the derivative of’ is a relation on any set of functions. We can define For example, let *f*(*x*) = *x*^{3}, *g*(*x*) = 3*x*^{2} and *h*(*x*) = 2*e*^{x}. Then *g* *f* because *g* is the derivative of *f*, and *h* *h* because *h* is the derivative of *h*, but *f* *g* because *f* is not the derivative of *g*.4. On , we can define a relation that is, *z*_{1} is related to *z*_{2} if the distance between *z*_{1} and *z*_{2} in the complex plane is less than or equal to 4. For example, 1 + *i* 2 − *i* because but 1 + *i* 3+ 5*i* because 4.2 Equivalence relationsOur formal definition of an equivalence relation involves three key properties. A relation that has these three properties partitions the set on which the relation is defined, as we show later in this subsection.The reflexive, symmetric and transitive properties are independent, in the sense that relations exist with every combination of these properties. (However, relations which are symmetric and transitive but not reflexive are usually somewhat contrived.)If a relation is symmetric, then ‘*x* is related to *y*’ means the same as ‘*y* is related to *x*’, and we can use either phrase, or simply say ‘*x* and *y* are related’; we can write either *x* *y* or *y* *x*.We now consider the examples in the previous subsection to see whether they satisfy any or all of the three properties.Examples1. The relation ‘is equal to’ on is reflexive, symmetric and transitive.It is reflexive since, for all *x* , *x* = *x*.It is symmetric since, for all *x*, *y* , if *x* = *y*, then *y* = *x*.It is transitive since, for all *x*, *y*, *z* , if *x* = *y* and *y* = *z*, then *x* = *z*.Hence this relation is an equivalence relation.2. The relation ‘is less than’ on is neither reflexive (since it is not true that *x* < *x* for all *x* ) nor symmetric (since, if *x* < *y*, then it does not follow that *y* < *x*), but it is transitive, since, if *x* < *y* and *y* < *z*, then *x* < *z*.3. The relation ‘is the derivative of ’ on a set of functions has none of the reflexive, symmetric and transitive properties.4. The relation defined on by is reflexive, since |*z* − *z*| = 0 ≤ 4 for all *z* . It is also symmetric, since, if |*z*_{1} − *z*_{2}| ≤ 4, then |*z*_{2} − *z*_{1}| = |*z*_{1} − *z*_{2}| ≤ 4. However, it is not transitive. The counter-example *z*_{1} = 0, *z*_{2} = 4, *z*_{3} = 4 + *i* shows that property E3 fails:So only the first of the four examples above is an equivalence relation.Example 17Prove that the relation defined on by is an equivalence relation on .SolutionWe show that properties E1, E2 and E3 hold.Hence this relation is an equivalence relation.Exercise 57For each set *A* and given relation, decide whether the relation has the reflexive, symmetric and transitive properties, and thus whether it is an equivalence relation. For each property, either prove that it holds or give a counter-example to show that it does not hold.(a) *A* = ; *x* *y* if *x* − *y* is odd.(b) *A* = ; *x* *y* if *x* − *y* is even.(c) *A* is the set of all lines in the plane; ℓ_{1} ℓ_{2} if the lines ℓ_{1} and ℓ_{2} are parallel. (Take the definition of parallel to be ‘in the same direction as’.)(d) *A* = ; *z*_{1} *z*_{2} if *z*_{1} − *z*_{2} is real.Solution(a) This relation is not reflexive; for example, 2 2 since 2 − 2 = 0 which is not odd.The relation is symmetric; if *x* *y* then *x* − *y* is odd, so *y* − *x* = − (*x* − *y*) is also odd, so *y* *x*.The relation is not transitive; for example, 5 2 since 5 − 2 is odd, 2 1 since 2 − 1 is odd, but 5 1 since 5 − 1 is even.The relation is not an equivalence relation.(b) This relation is reflexive, since *x* − *x* = 0, which is even, for all *x* .It is symmetric, since if *x* *y*, then *x* − *y* is even, so *y* − *x* = − (*x* − *y*) is even, so *y* *x*.It is transitive, since if *x* *y* and *y* *z*, then *x* − *y* is even and *y* − *z* is even, so *x* − *z* = *x* − *y* + *y* − *z* is the sum of two even numbers, so is even, so *x* *z*.The relation is an equivalence relation.(c) Any line ℓ is parallel to itself, so this relation is reflexive.It is symmetric, since if ℓ_{1} is parallel to ℓ_{2}, then ℓ_{2} is parallel to ℓ_{1}.It is transitive, since if ℓ_{1} is parallel to ℓ_{2} and ℓ_{2} is parallel to ℓ_{3}, then ℓ_{1} is parallel to ℓ_{3}.The relation is an equivalence relation.(d) This relation is reflexive, since for all *z* we have *z* − *z* = 0, which is real, so *z* *z*.It is symmetric, since if *z*_{1} − *z*_{2} is real, then *z*_{2} − *z*_{1} = − (*z*_{1} − *z*_{2}) is also real.It is transitive because, if *z*_{1} − *z*_{2} is real and *z*_{2} − *z*_{3} is real, then *z*_{1} − *z*_{3} = *z*_{1} − *z*_{2} + *z*_{2} − *z*_{3} is the sum of two real numbers and so is real.The relation is an equivalence relation.At the beginning of this section we stated that we were looking at a method of classifying objects in a set by *partitioning* them.DefinitionA collection of non-empty subsets of a set is a **partition** of the set if every two subsets in the collection are disjoint and the union of all the subsets in the collection is the whole set.(Two sets are *disjoint* if they have no elements in common.)We now show why an equivalence relation partitions the set on which the relation is defined. First we need another definition.DefinitionLet be an equivalence relation defined on a set *X*; then the **equivalence class** of *x* *X*, denoted by *x*, is the set Thus *x* is the set of all elements in *X* related to *x*.Example 18Find the equivalence classes for the following equivalence relations:(a) ‘is equal to’ on a set of real numbers;(b) the relation of Example 17: *z*_{1} *z*_{2} if |*z*_{1}| = |*z*_{2}|, on the set .Solution(a) Since *x* = *y* only if *y* is the same real number as *x*, the equivalence class of the real number *x* contains only the number *x* itself. So here each element lies in a single-element equivalence class.(b) The equivalence class of a particular complex number *z*_{0}, say, is *z*_{0} = {*z* : *z* *z*_{0}}. Now *z* *z*_{0} means that |*z*| = |*z*_{0}|, so *z*_{0} = {*z* : |*z*| = |*z*_{0}|}. Hence the equivalence class of *z*_{0} is the set of all complex numbers with the same modulus as *z*_{0}. If |*z*_{0}| = *r*, say, then *z*_{0} = {*z* : |*z*| = *r*}; this set forms the circle with centre 0 and radius *r* in the complex plane. Hence the equivalence classes for this relation are circles with centre 0. (The origin is an equivalence class containing just the complex number 0 + 0*i*; it can be thought of as a circle of radius 0.) Theorem 7The equivalence classes associated with an equivalence relation on a set *X* have the following properties.(a) Each *x* *X* is in an equivalence class.(b) For all *x*, *y* *X*, the equivalence classes *x* and *y* are either equal or disjoint.(Equivalence classes (being sets) are equal if they have exactly the same elements.)Thus the equivalence classes form a partition of *X*.**Proof**Let be an equivalence relation on a set *X*.(a) Let *x* *X*. The relation is reflexive, so *x* *x*, and hence *x* belongs to the equivalence class *x*.(b) Let *x* and *y* be equivalence classes with at least one element *a* in common.Then, since *x* and *y* are not disjoint, we have to prove that they are equal.First we show that *x* *y*. Suppose that *b* *x*; we have to show that *b* *y*. Since *a* *y*, *a* *x* and *b* *x*, we have respectively. (By definition, *a* *x* implies that *x* *a*, but since is symmetric, this means the same as *a* *x*.)Since is transitive, the first two statements of statements (4.1) together imply that *y* *x*, and this, together with the third statement of statements (4.1), implies that *y* *b*. Hence *b* *y*.This shows that *x* *y*. We can show similarly that *y* *x* (we interchange the roles of *x* and *y* in the proof that *x* *y*). Hence *x* = *y*.As an illustration of Theorem 7, consider the equivalence relation in Example 18(b). We saw that the equivalence classes of this relation are of the form where *r* . That is, this relation partitions the complex plane into concentric circles with centre the origin.(Here we think of the class containing the origin alone as a circle of radius 0.)It follows from Theorem 7 that if two elements *x* and *y* are related by an equivalence relation, then *x* = *y*. Thus, in general, there is more than one way to denote each equivalence class using the notation : a class can be denoted by *x* where *x* is any one of its elements. It is sometimes useful to choose a particular element *x* in each equivalence class and denote the class by *x*. The element *x* that we choose is called a *representative* of the class.For example, is one of the equivalence classes of the equivalence relation in Example 18(b). This class contains 4, −4*i* and 2 + 2*i*, for example, so we could denote it by any of 4, −4*i* or 2+2*i*. We might decide to choose the representative 4 and denote the class by 4. In general, the equivalence class Exercise 58Determine the equivalence classes for the equivalence relations in Exercise 57(b), (c) and (d).SolutionFor the equivalence relation in Exercise 57(b), the equivalence class *m* of a particular integer *m* is the set of integers that differ from *m* by an even integer. That is,In particular,These are the only equivalence classes (since the equivalence class of any other integer is equal to one of these two). So the equivalence classes are the even integers and the odd integers.For the equivalence relation in Exercise 57(c), the equivalence class of a particular line is the set of all lines that are parallel to that line. So each equivalence class consists of all lines with a particular gradient. For example, all vertical lines form one equivalence class, and all lines with gradient 3 form another.For the equivalence relation in Exercise 57(d), the equivalence class of a particular complex number *z*_{0} is If *z*_{0} = *x*_{0} + *iy*_{0} and *z* = *x* + *iy*, then which is a real number if and only if *y* = *y*_{0}. Thus *z*_{0} consists of all complex numbers with the same imaginary part as *z*_{0}. So each equivalence class consists of all complex numbers with a particular imaginary part. For example, all complex numbers of the form *x* + 3*i*, where *x* , form one equivalence class, and all complex numbers of the form *x* − *i*, where *x* , form another. The equivalence classes thus form horizontal lines in the complex plane.*Remark* These solutions are quite detailed, but you may find that you can determine equivalence classes with less working as you become more familiar with them.The solutions to Exercise 57(b) and Exercise 58 show that the relation on given by is an equivalence relation, with equivalence classes and that is, the even integers form one class and the odd integers the other. You have already met this idea in Section 3: the relation ‘*x* *y* if *x* − *y* is even’ is congruence modulo 2.For any *n*, congruence modulo *n* on , given by is an equivalence relation; the first three properties given in Theorem 5 are the reflexive, symmetric and transitive properties. The equivalence classes for this relation are the sets The representatives that we have used to denote the classes are 0, 1, 2, …, *n* − 1, which are the elements of _{n}. Thus _{n} is a *set of representatives* of the equivalence classes of congruence modulo *n*; that is, each equivalence class has exactly one representative in the set _{n}. The definitions of the modular operations +_{n} and ×_{n} can be rephrased using the idea of equivalence classes as follows: for all *a*, *b* _{n}, For example, in _{5},because 3 + 4 = 7 and the equivalence class 7 of congruence modulo 5 contains the element 2 of _{5}.So far, we have taken congruences only on , but it is possible to take congruences also on , and the modulus does not need to be an integer.Example 19Show that the relation defined on byis an equivalence relation, and describe the equivalence classes.SolutionWe show that properties E1, E2 and E3 hold.The equivalence class *r* of any real number *r* is the set of all real numbers related to *r* by ; that is, *r* is the set of all real numbers that differ from *r* by a multiple of 2. So The equivalence relation in Example 19 is congruence modulo 2. For example, and we write because We have seen that congruence modulo *n* on corresponds to modular arithmetic on _{n}, which is a set of representatives of the equivalence classes of congruence modulo *n*. In a similar way, congruence modulo 2 on corresponds to modular arithmetic on a set of representatives of the equivalence classes of congruence modulo 2. A suitable set of representatives is the interval (−, ], since every equivalence class has exactly one representative in this interval. (Other intervals can be used, for example [0, 2), but (−, ] is useful as it corresponds to our definition of the principal argument of a complex number.) We define modular operations and on the interval (−, ] as follows: for all *x*, *y* (−, ],For example,(
and
contains
)
.This is effectively what we do when we take the *principal* argument of a complex number arising from some calculation.Arithmetic modulo 2 on the interval (−, ] gives us a concise way to express some results about complex numbers. For example, we saw earlier that, if *z*_{1} and *z*_{2} are any two complex numbers, then Arg *z*_{1} + Arg *z*_{2} is an argument of *z*_{1}*z*_{2}, but is not necessarily the principal argument. The principal argument is Arg *z*_{1} Arg *z*_{2}, so we can now state that (Recall that Arg *z* denotes the principal argument of *z*.)4.3 Further exercisesExercise 59Let be the relation defined on byGive counter-examples to show that is not reflexive, symmetric or transitive.Solution is not reflexive because, for example, 1 1 since 2 × 1 − 1 = 1 is not divisible by 7. is not symmetric because, for example, 5 3 since 2 × 5 − 3 = 7 which is divisible by 7, but 3 5 since 2 × 3 − 5 = 1 which is not divisible by 7. is not transitive because, for example, 5 3 and 3 6 since 2 × 5 − 3 = 7 and 2 × 3 − 6 = 0 which are both divisible by 7, but 5 6 since 2 × 5 − 6 = 4 which is not divisible by 7.Exercise 60Let *A* be the set of all functions with domain and codomain , and let be the relation defined on *A* byShow that this is an equivalence relation and describe the equivalence classes.SolutionWe show that properties E1, E2 and E3 hold.E1 For any function *f* : → , *f*(0) = *f*(0), so *f* *f* and hence the relation is reflexive.E2 If *f* *g* so that *f*(0) = *g*(0), then *g*(0) = *f*(0), so *g* *f* and hence the relation is symmetric.E3 If *f* *g* and *g* *h* so that *f*(0) = *g*(0) and *g*(0) = *h*(0), then *f*(0) = *h*(0), so *f* *h* and hence the relation is transitive.Therefore this is an equivalence relation.Each equivalence class consists of all functions in *A* that take a particular value at 0; that is, each equivalence class is of the form for some *r* .Exercise 61Let be the relation defined on by where Show that this is an equivalence relation and describe the equivalence classes.SolutionWe show that properties E1, E2 and E3 hold.E1 Let *z* = *x* + *iy* . Then so *z* *z*. Hence the relation is reflexive.E2 Let *z*_{1} = *x*_{1} + *iy*_{1} and *z*_{2} = *x*_{2} + *iy*_{2} be elements of . Suppose that *z*_{1} *z*_{2} so that *x*_{1} − *x*_{2} = 5(*y*_{1} − *y*_{2}). Then so *z*_{2} *z*_{1}. Hence the relation is symmetric.E3 Let *z*_{3} = *x*_{3} + *iy*_{3}, and suppose that *z*_{1} *z*_{2} and *z*_{2} *z*_{3} so that and Then so *z*_{1} *z*_{3}. Hence the relation is transitive.Therefore this is an equivalence relation.Two complex numbers *x*_{1} + *iy*_{1} and *x*_{2} + *iy*_{2} are related by this relation if that is, if Hence the equivalence classes are the lines *x* − 5*y* = *r*, for each real number *r*; that is, the lines with gradient
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