# 3.2 Relationship between complex numbers and points in the plane

We have seen in Section 2.2 that the complex number system is obtained by defining arithmetic operations on the set **R** × **R**. We also know that elements of **R** × **R** can be represented as points in a plane. It seems reasonable to ask what insight can be obtained by representing complex numbers as points of the plane.

One obvious way of making the representation is to use Cartesian coordinates. Then the number *x* + *iy* is represented by the point with coordinates (*x, y*). Thus if a point *P* represents a complex number *z* then its *x* coordinate is the real part of *z* and its *y* coordinate is the imaginary part of *z* (hence the choice of notation *z* = *x* + *iy* made in Section 2.2).

The horizontal axis is called the **real axis** and the vertical axis is called the **imaginary axis**.

This way of representing complex numbers by points in the plane is used so often that we shall use phrases such as “the point *x* + *iy*” and “the point *z*” as convenient shorthands for “the point representing *z*” and we shall speak of the **complex plane**. Figure 1 shows several points in the complex plane.

As we saw in SAQ 11 in Section 2.4 (and it is fairly obvious anyway), the geometric representation can be taken a stage further. Complex numbers also add in the same way as number pairs regarded as vectors and so they add according to the parallelogram law (Figure 2).

Now points in the plane can also be represented by polar coordinates and this suggests another way of specifying a complex number.

If you have forgotten about polar coordinates, a short revision will not come amiss. When we use polar coordinates – instead of specifying a point in a plane by the distance from two lines, as in Cartesian coordinates – we specify a point *P* in the plane by its distance (*r*) from a fixed point *O* (the origin) and the angle (*θ*) that *OP* makes with a fixed half-line (the initial line) which has one end at *O* (Figure 3). (At present we consider only points *P* different from *O*.)

As is usual in mathematics, the angle *θ* is measured in an anticlockwise direction from the initial line (Figure 4.).

One snag that does arise with polar coordinates is that they do *not* give a one-one correspondence with the points of the plane. Thus a point with polar coordinates (*r*, *θ*) also has polar coordinates (*r*, *θ* + 2*k*) where *k* is any integer. But polar coordinates (*r*, *θ*) do specify a *unique* complex number, and from the relationship between polar coordinates and Cartesian coordinates we know that this number is *r* cos *θ* + *ir* sin *θ* (Figure 5). So whether we use polar coordinates or Cartesian coordinates, a point in the plane specifies a unique complex number.

It is often useful to represent a complex number in terms of the polar coordinates of the corresponding point as in the following definition.

## Definition

A **polar form** of the non-zero complex number

where

The positive number *r* is called the **modulus** of *z* and *θ* is **an argument** of *z*. We denote the modulus of *z* by |*z*|, and so For the special case of *z* = 0, we define |*z*| = 0, but we do not need to define what is meant by an argument of 0.

As we know, there will be some difficulty in specifying arguments. This is manifested by the fact that the relations specify *θ* only to within an integer multiple of 2. (Notice that the equation is even less specific.)

Given any non-zero complex number, *z*, an argument can be any element from a whole set of numbers. For example, an argument of 1 + *i* can be any element from . Nevertheless, any one element from this set is sufficient to identify the set and whatever the number *z*(≠0) there is only one of its arguments in the interval (−, ].

## Definition

If *θ* is an argument of *z* and θ(− ].*θ* is the **principal value of the argument** of *z*, and we write *θ* = Arg *z*. (Note the capital A.)

Thus, for example, (Figure 6).

It would have been just as reasonable to have chosen the interval [0.2) or (0.2] or [−, ) for Arg *z*, but we must settle for one option – we happen to have chosen (− ].

The polar form of a complex number can be very convenient and so we investigate now the effect of the arithmetic operations on the polar forms of complex numbers.

Addition and subtraction of complex numbers given in polar form is not a pleasant affair if the answers are also required to be in polar form. If

and

then

and that is about as far as you can get unless you are a wizard at trigonometry. (The polar form would be *p* cos *ψ* + *ip* sin *ψ* where *p* and *ψ* are expressions in *r*, *ρ*, *θ*, and *φ*.) Of course, in any particular case you can work out the polar form of the answer, but the general result is not worth working out, let alone remembering.

On the other hand, multiplication is easy because

Thus

and *an* argument of *zw* can be obtained just by adding arguments of *z* and *w*. If we want to work with principal values, there is one minor difficulty: even if *θ* and *φ* are in (−, ], when we add *θ* and *φ* we might well get an answer outside (−, ]. So if we want to work with principal values we shall have to make sure to get the answer back into (−, ]. Thus although Arg *z* + Arg *w* is an argument of *zw* we *may* have to add or subtract 2 to obtain Arg(*zw*).

One possible snag that may have occurred to you is this: if *θ* and *φ* are arguments of *z* and *w* respectively, is *θ* + *φ* an argument of *zw whatever* *θ* and *φ* we choose? Well, the relation where *θ*_{1} is related to *θ*_{2} if *θ*_{1} = *θ*_{2} + 2*k*, *k* an integer, is an equivalence relation on **R** and all the arguments of a complex number will constitute an equivalence class. We are asking whether we can combine equivalence classes (add arguments), in other words whether the equivalence relation is compatible with addition. Can we write

{arguments of *z*} + {arguments of *w*} = {arguments of *zw*}?

We can, of course, because if *θ*_{1} and *θ*_{2} are in the same equivalence class and *φ*_{1} and *φ*_{2} are in the same equivalence class, there are integers *k* and *m* such that

and so

that is, *θ*_{1} + *φ*_{1} is in the same equivalence class as *θ*_{2} + *φ*_{2}.

We can also show that if *α* is a fixed argument of *zw* then we can choose an argument of *z*, *θ* say, and an argument of *w*, *φ * say, such that *θ* + *φ* = *α*.

Let *α* = Arg *zw* + 2*n*, where *n* is a fixed integer. Certainly,

and so

So choose

where *p* + *q* = *k* + *n*.

We define **division of complex numbers** (as for real numbers) as the inverse of multiplication, and so it is no surprise that polar forms are also very suitable for division. If z^{−l} is the number such that

then we define *w* ÷ *z* by

As for real numbers, we write

We now ask how to calculate l/*z* when *z*(≠0) is given. If *θ* is an argument of *z* and *φ* an argument of 1/*z* then *θ* + *φ* is an argument of . Since 0 is an argument of 1, we must be able to choose *φ* so that

and so −*θ* is an argument of 1/z. In fact, unless *z* is a negative real number, we can use principal values without trouble:

(Note that a statement like *k* > 0 implies that *k* is real.)

If *z* = −*k*, *k* > 0, then Arg *z* = and Arg 1/*z* = .

What about |1/*z*|? We know that |1| = 1, and so

Thus

We can now interpret division quite easily in terms of polar forms:

and if *θ* and *φ* are arguments of *z* and *w* respectively, then *θ* − *φ* is an argument of *w/z*. More concisely,

To summarise: each of the arithmetic operations on **C** has a geometric interpretation. If *z* = *x* + *iy* is any complex number, corresponding to a point *P*, and if *w* = *u* + *iv* = *p*(cos φ + sinφ) then:

the effect of adding

*w*to*z*is to*translate**P*to the point (*x*+*u*,*y*+*v*); the effect of subtracting*w*from*z*is to*translate**P*to the point (*x*−*u*,*y*−*v*);-
the effect of multiplying

*z*by*w*is to*rotate**OP*through the angle*φ*and “*stretch*”*OP*by a factor*ρ*;the effect of dividing

*z*by w is to rotate*OP*through the angle −*φ*and “*stretch*”*OP*by a factor 1/*ρ*.

The fact that when multiplying complex numbers their arguments are added suggests an interesting relationship between numbers *z* and *w* for which

Multiplying these two numbers gives a number with a zero argument – a real number (Figure 7). What is more, the modulus of this number is |*z*| · |*w*| and this is just |*z*|^{2}. In polar form we see that if

The relationship between *z* and *w* is important and leads us to the following definition.

## Definition

If *z* = *x* + *iy*, the number *x* − *iy* is the **conjugate** of *z*, and we write

(Note that many authors write *z** for .)

We have just seen that has the important property that .

This result is useful for dealing with quotients in Cartesian form, for an expression like

could be simplified without resorting to polar form if we could make the denominator real. We now know how to do that. Since (*c*−*id*)/(*c*−*id*) = 1 , we can write

The details of this expression are *not* worth remembering, but the technique is: multiply numerator and denominator by the conjugate of the denominator. It is often useful to remember that a denominator *z* becomes |*z*|^{2} after the process.

Perhaps the most impressive payoff from writing complex numbers in polar form arises in the problem of solving an equation of the form

where *a* is a given non-zero complex number. Now it is easy to deduce from our knowledge of multiplication of polar forms that

and that if *θ* is an argument of *z* then *n**θ* is an argument of *z*^{n}. (See SAQ 21 in Section 3.4.) So if *z*^{n} = *a*, *z* must be a number whose modulus is an *n*th root of *a* and which has an argument which is 1/*n* times an argument of *a*.

## Example 1

Find a value of *z* such that *z*^{4} = 16*i*.

### Answer

### Solution

Since |16*i*| = 16, we have

An argument of 16*i* is , so an argument of *z* is , because . So is a possible value of *z*. But and this is also an argument of *a*, so

is another possible value of *z*.

This example clearly suggests that we need a *systematic* method of finding *all* the solutions of the equation *z*^{n} = *a*, *a* ≠ 0. We can see how to do this by considering the simpler problem of finding all the solutions of *w*^{n} = 1. Since |1| = 1, we must have and if *φ* is an argument of *w*, *n**φ* must be a multiple of *2* because 0 is an argument of 1. In other words, |w| = 1 and any of the values can be an argument of *w* – each leading to a different value of *w*. (If *k* takes integer values beyond *n*, we get repetitions of the values of

that have already been obtained.) Thus in the set of complex numbers, the number 1 has *n* *n*th roots:

The points representing the roots lie equally spaced around the unit circle. Figure 8 shows the roots for *n* = 8.

Now let us get back to the original problem: *z*^{n} = *a*, *a* ≠ 0. We do not seem to have any difficulty in finding one solution of this equation: the problem is how to find the whole set of solutions. Let us suppose we have one solution, *z*_{0}, so that *z*_{0}^{n} = *a*. We now use the solution to the simplified problem, *w ^{n}* = 1. Suppose

*w*is a solution of this problem, then (

*wz*

_{0})

^{n}=

*w*

^{n}

*z*

_{0}

^{n}= 1

*a*=

*a*, and so

*wz*

_{0}is a solution to the original problem. But there are

*n*possibilities for

*w*and so there are

*n*possibilities for

*wz*

_{0}(one of which is z

_{0}because one of the values of

*w*is 1). So knowing the solutions of

*w*= 1 (which is easy) and just one solution of

^{n}*z*= a (which is not too bad) we can generate

^{n}*n*solutions of

*z*=

^{n}*a*. The question is: “Have we now got them all?”. It would be most surprising if we had not, but we still have to check – or rather

*you*have to check, because that is SAQ 22 in Section 3.4.

In solving the equation *z*^{n} = *a* in the set of complex numbers we are extending the notion of roots of a real number and so we make the following definition.

## Definition

The solutions of *z*^{n} = *a* are called *n*th **roots** of *a*.

We have proved the following theorem (taking SAQ 22 of Section 3.4 for granted).

## Theorem 1

Every non-zero complex number has exactly *n* distinct *n*th roots.

We find the *n* *n*th roots of *a* ≠ 0 by using the following technique.

**Technique**

If then one solution of *z*^{n} = *a* is

and all the others are obtained by multiplying *z*_{0} in turn by each of the *n*th roots of *w*^{n} = 1.

Now the effect of multiplying each of the *n* *n*th roots of 1 by *z*_{0} will be to multiply each of their moduli (all equal to 1) by |*a*|^{1/n} and add *α*/*n* to each of their arguments. In other words, the *n*th roots of the complex number *a* are obtained from the *n*th roots of 1 by expanding (or contracting) the unit circle by |*a*|^{1/n} and rotating by (Arg *a*)/*n*. The *n*th roots of *a* will be equally spaced around a circle of radius |*a*|^{1/n}. The roots are

Figure 9 illustrates this for *n* = 8 and |a| > 1.

## Example 2

To find the solutions of *z*^{4} = 1 + *i*, the fourth roots of 1 + *i*, we proceed as follows.

Thus one root is and the others can be found because the four of them are equally spaced around the circle of radius (Figure 10).