# COMPLEX NUMBERS OPERATIONS WORKSHEET

## About "Complex Numbers Operations Worksheet"

Complex Numbers Operations Worksheet :

Worksheet given in this section is much useful to the students who would like to practice problems on complex numbers and operations.

Before look at the worksheet, if you would like to know the stuff related to complex numbers and operations,

## Complex Numbers Operations Worksheet - Problems

Problem 1 :

Solve the following quadratic equation using square root :

x2  =  36

Problem 2 :

Solve the following quadratic equation using square root :

x2  =  -25

Problem 3 :

Find the sum of the following two complex numbers.

(3 - 4i) and (-8 + 6i)

Problem 4 :

Find the difference of the following two complex numbers.

(5 + 9i) and (3 - 4i)

Problem 5 :

Find the product in the form a + bi :

-3.5i(7 - 8i)

Problem 6 :

Find the product in the form a + bi :

(5 + 3i)(5 - 3i)

Problem 7 :

Write the following complex number in a + bi form :

5 / (2 - i)

Problem 8 :

Factor p2 + q2.

Problem 9 :

Factor 16p2 + 4.

Problem 10 :

Solve the following quadratic equation using factoring :

x2 + 16  =  0

## Complex Numbers Operations Worksheet - Solutions

Problem 1 :

Solve the following quadratic equation using square root :

x2  =  36

Solution :

In the given quadratic equation, x term is missing. So we can solve the equation using square root.

x2  =  36

Take square root on each side.

x  =  ±√36

x  =  ± 6

The solutions of the equation x2  =  36 are -6 and 6.

Problem 2 :

Solve the following quadratic equation using square root :

x2  =  -25

Solution :

In the given quadratic equation, x term is missing. So we can solve the equation using square root.

x2  =  -25

Take square root on each side.

x  =  ± √-25

x  =  ± √25-1

x  =  ± 5i

The solutions of the equation x2 = -25 are not real numbers but are part of a number system called the complex numbers. The number -1 is called the imaginary unit i. Replacing -1 with i allows us to write the solutions to the equation x2  =  -25 as 5i and -5i.

Problem 3 :

Find the sum of the following two complex numbers.

(3 - 4i) and (-8 + 6i)

Solution :

When adding two complex numbers in the form a + bi, combine the real parts and then combine the imaginary parts. The sum will include both a real and imaginary part and can be written in the form a + bi.

(3 - 4i) + (-8 + 6i)  =  [3 + (-8)] + [-4i + 6i]

=  [3 - 8] + [2i]

=  - 5 + 2i

Problem 4 :

Find the difference of the following two complex numbers.

(5 + 9i) and (3 - 4i)

Solution :

When subtracting two complex numbers in the form a + bi, combine the real parts and then combine the imaginary parts. The difference will include both a real and imaginary part and can be written in the form a + bi.

(5 + 9i) - (3 - 4i)  =  [5 - 3] + [9i - (-4i)]

=   [2] + [9i + 4i]

=   2 + 13i

Problem 5 :

Find the product in the form a + bi :

-3.5i(7 - 8i)

Solution :

Use the distributive property.

-3.5i(7 - 8i)  =  -3.5i(7) - 3.5i(-8i)

Multiply.

=  -24.5i + 28i2

Simplify using the definition i2.

=  -24.5i + 28(-1)

=  -24.5i - 28

Write in the form a + bi.

=  -28 - 24.5i

Problem 6 :

Find the product in the form a + bi :

(5 + 3i)(5 - 3i)

Solution :

Use the distributive property.

(5 + 3i)(5 - 3i)  =  5(5 - 3i) + 3i(5 - 3i)

Multiply.

=  25 - 15i + 15i - 9i2

=  25 - 9i2

Simplify using the definition i2.

=  25 - 9(-1)

=  25 + 9

=  34

Problem 7 :

Write the following complex number in a + bi form :

5 / (2 - i)

Solution :

When the denominator has an imaginary component, we can create an equivalent fraction with a real denominator by multiplying by its complex conjugate.

Use the complex conjugate of the denominator to multiply by 1.

5 / (2 - i)  =  [5 / (2 - i)]  x  1

=  [5 / (2 - i)]  x  [(2 + i) / (2 + i)]

=  [5(2 + i)]  /  [(2 - i)(2 + i)]

Use the distributive property.

=  (10 + 5i)  /  (4 + 2i - 2i - i2)

=  (10 + 5i)  /  (4 - i2)

Simplify using the definition i2.

=  (10 + 5i)  /  [4 - (-1)]

=  (10 + 5i)  /  (4 + 1)

=  (10 + 5i)  /  5

=  10 / 5  +  5i /5

=  2 + i

Problem 8 :

Factor p2 + q2.

Solution :

Rewrite p2 + qas a difference of two squares :

p2 - (-q2)

We can think of (-q2) as

(-1)(q2)

Because -1  =  i2, we have

(-q2)  =  (-1)(q2)  =  (i2)(q2)  =  (iq)2

So, we have

p2 + q2  =  p2 - (-q2)

=  p2 - (iq)2

=  (p + iq)(p - iq)

Problem 9 :

Factor 16p2 + 4.

Solution :

Factor out the GCF :

16p2 + 4  =  4(4p2 + 1)

Rewrite as a difference of squares.

=  4(4p2 - i2)

Factor the difference of squares.

=  4(4p2 - i2)

=  4(4p + i)(4p - i)

The factors of 16p2 + 4 are 4, (4p + i) and (4p - i).

Problem 10 :

Solve the following quadratic equation using factoring :

x2 + 16  =  0

Solution :

Write the original equation :

x2 + 16  =  0

x2 + 42  =  0

Rewrite as a difference of squares.

x2 - (4i)2  =  0

(x + 4i)(x - 4i)  =  0

By Zero Product Property,

x + 4i  =  0,    x - 4i  =  0

x  =  -4i,    x  =  4i

The solutions are -4i and 4i.

After having gone through the stuff given above, we hope that the students would have understood, "Complex Numbers and Operations".

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