**Complex Numbers and Operations :**

Complex numbers are numbers that can be written in the form a + b*i*, 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,* i*^{2} = -1.

An imaginary number is any number, b*i*, 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)

**Example 1 :**

Solve the following quadratic equation using square root :

x^{2} = 36

**Solution : **

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

x^{2} = 36

Take square root on each side.

x = ±√36

x = ± 6

The solutions of the equation x^{2} = 36 are -6 and 6.

**Example 2 :**

Solve the following quadratic equation using square root :

x^{2} = -25

**Solution : **

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

x^{2} = -25

Take square root on each side.

x = ± √-25

x = ± √25√-1

x = ± 5i

The solutions of the equation x^{2} = -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 x^{2} = -25 as 5i and -5i.

**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

**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 + 28i^{2}

Simplify using the definition i^{2}.

= -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 - 9i^{2}

= 25 - 9i^{2}

Simplify using the definition i^{2}.

= 25 - 9(-1)

= 25 + 9

= 34

**Note :**

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

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 - 4i^{2}

= 9 - 4(-1)

= 9 + 4

= 13

**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 - i^{2})

= (10 + 5i) / (4 - i^{2})

Simplify using the definition i^{2}.

= (10 + 5i) / [4 - (-1)]

= (10 + 5i) / (4 + 1)

= (10 + 5i) / 5

= 10 / 5 + 5i /5

= 2 + i

**Example 1 : **

Factor p^{2} + q^{2}.

**Solution : **

Rewrite p^{2} + q^{2 }as a difference of two squares :

p^{2} - (-q^{2})

We can think of (-q^{2}) as

(-1)(q^{2})

Because -1 = i^{2}, we have

(-q^{2}) = (-1)(q^{2}) = (i^{2})(q^{2}) = (iq)^{2}

So, we have

p^{2} + q^{2} = p^{2} - (-q^{2})

= p^{2} - (iq)^{2}

= (p + iq)(p - iq)

**Example 2 : **

Factor 16p^{2} + 4.

**Solution : **

Factor out the GCF :

16p^{2} + 4^{ }= 4(4p^{2} + 1)

Rewrite as a difference of squares.

= 4(4p^{2} - i^{2})

Factor the difference of squares.

= 4(4p^{2} - i^{2})

= 4(4p + i)(4p - i)

The factors of 16p^{2} + 4 are 4, (4p + i) and (4p - i).

**Example : **

Solve the following quadratic equation using factoring :

x^{2} + 16 = 0

**Solution : **

Write the original equation :

x^{2} + 16 = 0

x^{2} + 4^{2} = 0

Rewrite as a difference of squares.

x^{2} - (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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