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". 

Apart from the stuff given in this section if you need any other stuff in math, please use our google custom search here.

Widget is loading comments...