# 2 Complex numbers

## 2.11 Further exercises

### Exercise 28

Let *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 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) −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 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 + 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 31

Let *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)

### Exercise 32

Solve the equation *z*^{5} = −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 *z*^{5} = −32 are given by

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

Hence the solutions are

### Exercise 33

Solve the equation *z*^{3} + *z*^{2} − *z* + 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 *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 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 6*θ* and sin 6*θ* in terms of cos *θ* and sin *θ*.

### Solution

Hence

and

### Exercise 36

Use 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}.