Complex numbers
Complex numbers

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Complex numbers

3.4 Self-assessment questions and problems

SAQ 13

Find |z| and Arg z in each of the following cases.

Answer

SAQ 14

Express each of the following in the form a + ib where a and b are real.

  1.  
  2.  
  3.  
  4.  
  5.  
  6.  

Answer

SAQ 15

Find the modulus (r) and an argument (θ) for each of the following numbers.

Answer

(i)

(We asked you only for an argument: you could have given any one of , where k is an integer. An alternative way of expressing this is to write θ = /4 (mod 2). Similar remarks apply to the other parts.)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

SAQ 16

Express each of the following numbers in the form a + ib, where a and b are real numbers.

Answer

SAQ 17

Show that for any complex numbers z and w

Answer

(i)

(Note that we have taken z = x + iy; since this is the standard representation, we do not normally state it explicitly.)

(ii)

SAQ 18

Illustrate the following sets of points of the complex plane where a, b, c, d α and β are fixed real numbers with b > a, d > c, − < α < β < .

Answer

(i)

Figure 15
Figure 15

(ii)

Figure 16
Figure 16

(iii)

Figure 17
Figure 17

(iv)

Figure 18
Figure 18

(v)

Figure 19
Figure 19

(vi)

and this is zero if and only if

Thus x and y satisfy

which is equivalent to

The locus is a circle with centre and radius (Figure 20). The point z = 2 on this circle must be excluded from the locus, because (z + 3)/(z − 2) is not defined for z = 2.

Figure 20
Figure 20

When you are used to manipulating complex numbers you will be able to work more elegantly as follows. gives since is real. Thus

and so

(vii)

Figure 21
Figure 21

(viii)

Figure 22
Figure 22

(ix) 3Re z + 2 Imz < 6 corresponds to 3x + 2y < 6 and so the set is as indicated in Figure 23.

Figure 23
Figure 23

(x)

Figure 24
Figure 24

(The regions in Figures 15, 16 and 23 are referred to as “half-planes”. Figures 17 and 18 illustrate “strips”; Figure 19 shows a “rectangle”, and Figure 24 a “triangle”. Note also that in Figures 15 to 19 we have illustrated only the case a, b, c and d positive. In Figure 21 we have chosen α < −/2 and β  > /2. In Figure 22 we have chosen 0 < α < /2.)

Style for figures: It is difficult to develop a completely consistent style for figures. However, you should find the following guidelines useful. (i) Infinite (unbounded) sets are shown by squared-off shading as in Figures 15, 16, 21, 22 and 23. (ii) Sets which do not include their “boundary” lines have thin boundary lines (for example, Figures 15, 16, 17, 18, 21, 23 and 24). (iii) Thick boundary lines indicate that the set includes its boundary lines. (iv) Most other conventions are self-evident or obvious from the context.

SAQ 19

If n is a multiple of 4, show that the coefficients, ci in SAQ 7 of Section 2.4 satisfy

and

Identify the cases when the positive sign should be taken on the right-hand side in the second case. (There is a hint after the solutions.)

Answer

Substituting gives

Since is real if n is a multiple of 4, and so equating real and imaginary parts gives

Since , we have .

We have seen that if n is a multiple of 4 then (1 +i)n is real, that is Arg(1 + i)n is either 0 (in which case (1 + i)n is positive) or (in which case (1 +i)n is negative). But, since an argument of 1 + i is /4, an argument of (1 + i)n is n/4.

If n is an even multiple of 4, that is n = 2k × 4 for some integer k, an argument of (1 + i)n is 2k. In other words, Arg(1 + i)n = 0 and so (1 + i)n is positive.

If n is an odd multiple of 4, that is n = (2k + 1) × 4 for some integer k, an argument of (1 + i)n is (2k+ 1). In other words, Arg(1 + i)n =  and so (1 + i)n is negative.

Hint

Try substituting a suitable value for z in

SAQ 20

Let p(z) be a polynomial with real coefficients.

(i) Prove that .

(ii) If α is a root of p(z) show that is also a root.

Answer

(i)

By an extension of a result in SAQ 17(iii) Also, if ck is real, , and again from SAQ 17(ii), the conjugate of a sum is the sum of the conjugates and so

(ii) If α is a root of the polynomial then and so and thus from (i), ; in other words, is also a root.

SAQ 21

(i) Prove by mathematical induction that if n is any positive integer, then |zn| = |z|n.

(ii) Show that if n is any positive integer, then

This result is known as de Moivre's Theorem. Abraham de Moivre (1667–1754) was born in France, but came to England when still a small boy. His interest in Mathematics started when he came across a copy of Newton's Principia by chance. He is best known for the theorem stated above, and for calculating the various quadratic factors of x2n−2pxn + 1, and is also remembered as one of the founders of probability theory.

(iii) Show that if n is any positive integer, then

This extends de Moivre's Theorem to any integer n.

(iv) We have seen in the text that de Moivre's Theorem extends to the form: if m is a positive integer, one of the values of (cos θ + i sin θ)1/m is cos (θ/m) + i sin (θ/m). Extend de Moivre's Theorem to any rational index.

Answer

(i) The result is trivially true for n = 1.

If the result is true for n = k, that is |zk| = |z|k, then

The result follows by mathematical induction.

(ii) The result is trivially true for n = 1. If the result is true for n = k, then

from the ordinary properties of multiplication of two complex numbers. The result follows by mathematical induction.

(iii) This follows from (ii) by noting that

Hence

(iv) Let r be any rational number. Then r = n/m where n is an integer and m is a positive integer. Since m is a positive integer, one of the values of (cosθ + isinθ )1/m is cos (θ/m) + i sin (θ/m) , and so, from (ii) and (iii), one of the values of (cos θ + isin θ)n/mis cos (n θ/m) + i sin(nθ/m).

SAQ 22

Show that if z0 is a solution of zn = a, where a ≠ 0, then any other solution must be of the form z0w where w is a solution of wn = 1.

Answer

Since a ≠ 0, we have z0 ≠ 0 and so any complex number z can be written in the form z0w, just by choosing . If z is a solution of zn = a, then

(z0w)n = a

and so

z0nwn = a.

But z0n = a and so we must have wn = 1.

SAQ 23

(i) Find the fifth roots of −32.

(ii) Find the fourth roots of −16.

Answer

(i) Since −32  = 32 [cos (-)  + isin (-)] , one fifth root of −32 is

Since the five fifth roots of −32 are equally spaced around the circle with centre at the origin and radius 2, the other roots are

(Note that k = 3 gives the real fifth root −2. The above method is the systematic one; we could, however, have made use of −2 as the one obvious fifth root of −32 from the start.)

(ii) Since −16 = 16[cos (−)  + i sin (−)]. , one fourth root of −16 is

Since the four fourth of −16 are equally spaced around the circle with centre at the origin and radius 2, the other roots are

SAQ 24

(i) Solve the equation z4 + 4z2 +  8 = 0.

(ii) Solve the equation z4 + 4iz2 +  8 = 0.

Answer

(i) Putting w = z2 gives w2 + 4w + 8  = 0,

and so

. So z can take the values of the two square roots of −2 + 2i and the two square roots of −2 −2i. Since

the two square roots of −2 + 2i are

and

Similarly, we can show that the two square roots of −2 −2i are

(Alternatively, we could use the result of SAQ 20 (ii) to find the last two roots, since is a polynomial with real coefficients.)

(ii) Putting w = z2 gives

and so

that is

Since , the two square roots of are

and

Since

and

Note that the four roots of do not occur in conjugate pairs: the coefficient of z2 is not real.

M332_1

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