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.

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)  −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.

Solution

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.