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 - i√2  ----> (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 + q² as 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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