Statement of a Problem

CHAPTER 12 SOLUTIONS TO EXERCISES ON COMPLEX NUMBERS

Exercise on 12.1

Solve the following equations in a + jb form:-

i) x2 + 4 = 0 ii) x2 + x + 1 = 0 iii) x2 + 6x + 11 = 0

iv) x3 – 1 = 0

Solutions

i) x2 + 4 = 0

We have

x2 = – 4 or x = ± = ± 2= ± 2j

ii) x2 + x + 1 = 0

Using the formula for the solution of a quadratic we have

x = = = – ± j

iii) x2 + 6x + 11 = 0

x = = = – 3 ± j

iv) x3 – 1 = 0

Here we first factorize the cubic to give

x3 – 1 = (x – 1)(x2 + x + 1) = 0

This gives solutions, from ii)

x = 1, – ± j

Exercise on 12.2

For the complex numbers z = 3 – j and w = 1 + 2j evaluate

i) 2z – 3w ii) zw iii) z2w 2 iv)

Solutions

i) 2z – 3w = 2(3 – j) – 3(1 + 2j) = 6 – 2j – 3 – 6j = 3 – 8j

ii) zw = (3 – j)(1 + 2j) = 3 – j + 6j – 2 j2 = 3 + 5j – 2(– 1) = 5 + 5j

on remembering that j2 = – 1

iii) Don't be so hasty with z2w 2 – look before you leap and notice that

z2w 2 = w(z )2 = (1 + 2j)= 10(1 + 2j) = 10 + 20j

iv) = = = =

on multiplying top and bottom by the conjugate of 1 + 2j

Exercises on 12.3

1. Plot the following complex numbers on the Argand plane and put them into polar form.

i) 1 ii) j iii) – 3j iv) 1 – j v) 2 + j vi) – 3 – 2j viii) – 3 + 2j

Solution

[pic]

i) 1 has modulus r = = 1 and as it lies on the positive real axis the principal value of its argument is 0 radians, and so

1 = 1 ((0)

ii) |j| = 1 and j lies on the imaginary axis so its argument is θ = . So

j = 1 (

iii) |– 3j| = 3 and the PV of the argument is θ = – , so

–3 j = 3 (

iv) |1 – j| = so

1 – j =

=

=

= (

So

1 – j = (

v) 2 + j = = ((θ) where tan θ =

vi) – 3 – 2j = = ((θ) where tan θ = π + tan–1

Here the complex number is in the third quadrant and we have chosen the angle in the range...