# 2 Complex numbers

## 2.11 Further exercises

### Exercise 28

Let z1 = 2 + 3i and z2 = 1 − 4i. Find z1 + z2, z1z2, z1z2, , , z1/z2 and 1/z1.

### Exercise 29

Draw 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)  −5i
(c)  −2 − 2i

### 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 = −5i, 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 −5i in terms of the principal argument is

(c)  Let z = x + iy = −2 − 2i, 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 − 2i in terms of the principal argument is

### Exercise 30

Express 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 + 2i.
(b)  The required form is x + iy, where

so the Cartesian form is 3i.
(c)  The required form is x + iy, where

and

so the Cartesian form is

### Exercise 31

Let z1 =i, z2 = −5i and z3 = −2 −2i. Use the solution to Exercise 29 to determine the following complex numbers in polar form in terms of the principal argument.

(a)  z1z2z3
(b)

### Solution

From the solution to Exercise 29, we have

(a)  Hence

using the principal argument.
(b)

### Exercise 32

Solve the equation z5 = −32, leaving your answers in polar form.

### Solution

Let 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 z5 = −32 are given by

for k = 0, 1, 2, 3, 4.

Hence the solutions are

### Exercise 33

Solve the equation z3 + z2z + 15 = 0, given that one solution is an integer.

### Solution

The 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 z2 −2z + 5 = 0 are given by

Hence the solutions of z3 + z2z + 15 = 0 are z = −3, z =1 + 2i and z = 1 − 2i.

### Exercise 34

Determine a polynomial of degree 4 whose roots are 3, −2, 2 − i and 2 + i.

### Solution

A suitable polynomial is

that is,

or

### Exercise 35

Use de Moivre's Theorem to obtain formulas for cos 6θ and sin 6θ in terms of cos θ and sin θ.

Hence

and

### Exercise 36

Use the definition of ez to express the following complex numbers in Cartesian form.

(a)
(b)
(c)

### Solution

(a)

(b)

(c)   = e−1(cos + isin) = −e−1.