Page | 1

Practice Test on Topic 18 Complex Numbers

Test on Topic 18Complex Numbers

1. Express the following as complex numbers a + bi

(a) (b) 2 25

(c) 2 + 8 (d) 4 + 9

(e) 4 + 16 (f) 36 + 25

2. Write each of the following as a complex number in the form a + bi

(a) 81 36 (b) (2i – 1)(3i – 2)

3. Write each of the following as a complex number in the form a + bi.

(a) (5 + 4i) + (3 + 2i)

(c) 2(3 + 2i) + 5(4 – 3i)

(b) (3 + i) – (7 + 2i)

(d) 4(3 + 4i) – 2(6 – 6i)

4. Write each of the following as a complex number in the form a + bi.

(a) 2i(3 + 4i)

(c) (2 + i)(4 – 3i)

(e) (2 + 3i)2

(b) 5i(3 + 4i)

(d) (5 + 4i)(5 – 4i)

(f)

5.

6.

The quadratic equation x2 – 2x + 6 = 0 has two solutions which are both complex numbers.Use the quadratic formula to find these solutions.

The quadratic equation x2 – x + 4 = 0 has two solutions which are both complex numbers.Use the quadratic formula to find these solutions.

Practice Test on Topic 18 Complex Numbers

Complex Numbers Solutions

1. Express the following as complex numbers a + bi

(a) = 0 + 3i

(b) 2 25 = 2 – 5i

(c) 2 + 8 = 2i 2 2i = 0 3 2i

(d) 4 + 9 = 4+ 9 1 = 4+ 9 1 = 4 + 3i

(e) 4 + 16 = 4 1 + 16 1===

4 1 + 16 12i + 4i6i

(f) 36 + 25 = 6i + 5i = 0 + 11i

2. Write each of the following as a complex number in the form a + bi

(b) 81 36 = 9i +6i = 15i

(b) (2i – 1)(3i – 2) = 6i2 – 4i – 3i + 2 = – 6 – 7i + 2 = – 4 – 7i

3. Write each of the following as a complex number in the form a + bi.

(a) (5 + 4i) + (3 + 2i)= 5 + 3 + 4i + 2i = 8 + 6i

(c) (3 + i) – (7 + 2i) = 3 + i – 7 – 2i = 3 – 7 + i– 2i = – 4 – i

(c) 2(3 + 2i) + 5(4 – 3i)

(d) 4(3 + 4i) – 2(6 – 6i)

=

=

6 + 6i + 20 – 15i

12 + 16i – 12 + 12i

=

=

9 + 20 + 6i – 15i

12 – 12 + 16i + 12i

=

=

29 – 9i

28i

4. Write each of the following as a complex number in the form a + bi.

Page | 2

Practice Test on Topic 18 Complex Numbers

Page | 3

(a) 2i(3 + 4i) = 6i + 8i2 Using the distributive law===

6i + 8(–1)6i – 8– 8 + 6i

Using the property that i2 = – 1

Rearranging terms

(b) 5i(3 + 4i) = 15i + 20i2

===

Using the distributive law15i + 20(–1) Using the property that i2 = – 115i – 20– 20 + 15i

Rearranging terms

Using FOILUsing the property that i2 = – 1

Using FOILUsing the property that i2 = – 1

(c) (2 + i)(4 – 3i) ====

(d) (5 + 4i)(5 – 4i) ====

8 – 6i + 4i – 3i2

8 – 2i – 3(–1)8 – 2i + 311 – 2i

25 – 20i + 20i – 16i2

8 – 2(–1)8+210

(e) (2 + 3i)2 = (2 + 3i)(2 + 3i)====

4 + 6i + 6i + 9i2

4 + 12i + 9(–1)4 + 12i – 9– 5 + 12i

Using the distributive lawUsing the property that i2 = – 1

Adding and subtracting like terms

(f) First we convert all the numbers to be in terms of i

==

( - )

iUsing the distributive lawUsing the property that i2 = – 1

=

===

– 4 – 12i

Practice Test on Topic 18 Complex Numbers

Page | 4

5. The quadratic equation x2 – 2x + 6 = 0 has two solutions which are both complex numbers.Use the quadratic formula to find these solutions.

Solution: x2 – 2x + 6 = 0a=1b= – 2c=6

b2 – 4ac = (–2)2 – 4(1)(6) = 4 – 24 = – 20

So b 2 4ac = 20 = 20 i = 2 5 i

x = b b 2 4ac2a

= (4 ) 2 5 i2(1)

=4 2 5 i2

The two solutions are x =4 2 5 i2

= 2 – 5 i

and x =4 2 5 i2

= 2+ 5 i

There are 2 complex number solutions they are x = 2 –5 i and x = 2 + 5 i

6. The quadratic equation x2 – x + 4 = 0 has two solutions which are both complex numbers.Use the quadratic formula to find these solutions.

Solution: x2 – 2x + 2 = 0a=1 b= – 1 c=4

b2 – 4ac = (– 1)2 – 4 x 1 x (4) = 1 – 16 = – 15

So b 2 4ac = 15 = 15 1 = 15 1 = 15 i

=2 15i2

x

So x

or

x

=

=

=

b b 2 4ac2a

2 15i2

2 15i2

There are two complex numbers as solutions they are x =2 15i

2and x =

2 15i2