# COMPLEX NUMBERS AND OPERATIONS

## About "Complex Numbers and Operations"

Complex Numbers and Operations :

Complex numbers are numbers that can be written in the form a + bi, where a and b are real numbers and i is the square root of -1. They include all real and imaginary numbers, as well as the sum of real and imaginary numbers.

The imaginary unit i, is the principal square root of -1.

Then, i2  =  -1.

An imaginary number is any number, bi, where b is a non-zero real number and i is the square root of -1.

For example :

-6 + 4i  ----> (a  =  -6, b  =  4)

7 - i2  ----> (a  =  7, b  =  -2)

0.5i  ----> (a  =  0, b  =  0.5)

## Solving Quadratic Equations Using Square Roots

Example 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.

Example 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.

## Add and Subtract Complex Numbers

Example 1 :

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

Example 2 :

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

## Multiply Complex Numbers

Example 1 :

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

Example 2 :

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

Note :

In the product in example 2, there is no imaginary part.

## Complex Conjugate

Complex conjugate are complex numbers with equivalent real parts and opposite imaginary parts. Their product is a real number.

For example :

3 + 2i,  3 - 2i

-1 + i,  -1 - i

Product :

(3 + 2i)(3 - 2i)  =  3(3 - 2i) + 2i(3 - 2i)

=  9 - 6i + 6i - 4i2

=  9 - 4(-1)

=  9 + 4

=  13

## Simplify a Quotient with Complex Numbers

Example :

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

## Factor a Sum of Squares

Example 1 :

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)

Example 2 :

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).

## Solve a Quadratic Equation with Complex Solutions

Example :

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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