In this post we try to understand the basic of complex numbers, its basic principles, examples and method of calculation like addition, subtraction, multiplication and division.

The concepts are important for students of Grade 11 as its questions are directly asked in the exams but the principle of complex number are relevant for your higher studies or your professional career.

- Complex Number Concepts
- What are complex numbers?
- What is ” i ” in math expression z = a + ib?
- complex numbers with equal value
- Complex Number Algebra

**Complex Number Concepts**

Do you know what is the answer of \sqrt { -1 } ?

From what you have studied till now, you know that the square root of negative number is not possible.

For this type of case you need to understand the concept of complex number. In this post i will introduce you to principle of complex number and teach you how to perform basic math operation on these numbers

**What are complex numbers?**

Complex number is basically a combination of real and imaginary number.

Basically the numbers are represented as:

The number 5 +10i is a complex number which is made up of two elements:

Real element (here 5) and imaginary element (here 10i)

Generally we can represent the complex number in following form:**z = a + ib**

here**a= real element****ib = is the imaginary element****z is annotation of complex number**

**What is ” i ” in math expression z = a + ib?**

“i” is the imaginary unit in the expression. If you combine any real number with unit ” i “, the number become imaginary element.

For instance, in the example of “5 + 10i” which i have mentioned above, when the real number 10 is combined with element “i”, the expression “10i” becomes the imaginary element

**Value of “i”**

The number “i” is represented as square root of minus one

i=\sqrt { -1 }

Hence with the help of “i”, we can find the solution of square root of negative number which was not possible for us earlier.

**Power of “i”**

We already know that i= \sqrt{-1}

We will now try to understand the value of “i” with different powers

i^{2} =\ -1\\ \\
i^{3} =i\times i^{2} =-i \\ \\
i^{4} =\left( i^{2}\right)^{2} =( -1)^{2} =1 \\ \\
i^{5} =\ \left( i^{2}\right)^{2} \ i\ =i \\ \\
i^{6} =\left( i^{2}\right)^{3} =( -1)^{3} =-1 \\ \\

From observing the above values we can generalize the power of “i” as

i^{4k} =1\\ \\
i^{4k+1} =i\\ \\
i^{4k+2} =-1\\ \\
i^{4k+3} =-i \\ \\

Where k is integer value given in the question

Please try to remember above formula as it will help you solve questions of complex number in less time.

**Method to remember the value of “i” formula**

**complex numbers with equal value**

Let us suppose that we have been provided with two complex number

z1= a + ib

z2 =c + id

Both the complex number z1 and z2 are equal only if their real number and imaginary number are equal. i.e **a = c** and **b = d**

## **Complex Number Algebra**

In this post we will understand how to perform math operations like addition, subtraction, multiplication and division of complex numbers. The concept is simple and straightforward but you need to invest some time so that you get used to the process.

**Addition of complex number**

Addition of two complex number can be simply done by adding the real part and imaginary part separately.

Suppose there are two complex numbers z1 and z2 such that

z1 = a + ib

z2= c + id

Then the addition of complex number z1+z2 can be done as**z1 + z2= (a + c) + i (b+d)**

The process is just like simple arithmetic operation, just you have to keep the calculation of real and imaginary part separate from each other

**Example 01**

z1 = 5 + 10i

z2 = 9 +3i

Find z1 + z2 of the given complex numbers

**Solution**

(z1 + z2) = (5 + 9) + (10+3)i

(z1 + z2) = 14 + 13i

**Example 02**

z1= -4 + 6i

z2 = 4 + 7i

Find (z1 + z2) for the given set of complex numbers

**Solution**

z1 + z2 = (-4 + 4) + (6+7)i

z1 + z2 = 0+13i

z1 + z2 = 13i

I hope you have now understood the concept of addition of complex number, now we will look into some of the important properties associated with the addition of complex numbers

**Property of Addition of complex number**

**(a) Closure Law**

It states that the addition of two complex number will also a complex number

**(b) Commutative Law**

In addition of complex number, if the position of complex number is interchanged the final result will be the same

**(c) Associative Law**

In addition of three complex numbers, if we group different numbers for addition, the result will be the same

**(d) Additive inverse of complex number (-z)**

suppose **z = a + ib** is the given complex number.

Its additive inverse can be written as:**(-z) = -a – ib**

Its called an additive inverse because adding the inverse with the given complex number will result in 0.**z + (-z) =0**

i hope you have now understood the concept of addition of complex number. Now we will move on to other operations which are possible in this chapter.

**Subtraction of Complex Number**

Subtraction of two complex number is the difference of real number and imaginary number of the given complex numbers

Let the two complex numbers are **z1 = a + ib****z2 = c + id**

The difference of complex number (z1 – z2) is given as:**(z1 – z2) = (a + ib) – (c + id)(z1 – z2) = (a – c) + i(b – d)**

**Example 01**

z1 = 5 + 10i

z2 = 3 + 6i

Find (z1 – z2)

**Solution**

(z1 – z2) = (5 – 3) + (10-6)i

(z1 – z2) = 2 + 4i

**Example 02**

z1 = -4 + 4i

z2 = -3 + 6i

Find (z1 – z2)

**Solution**

(z1 – z2) = (-4 + 3) + (4 – 6)i

(z1 – z2) = – 1 – 2i

**Multiplication of complex number**

Suppose there are two complex number**z1 = a + ibz2 = c + id**

Let us multiply the two given complex number z1 * z2

z1 * z2 = (a + ib) * (c + id)

z1 * z2 = ac + iad +ibc + { i }^{ 2 }\quad bd

we know that { i }^{ 2 } = -1

z1 * z2 = ac + iad +ibc -bd

**z1 * z2 = (ac – bd) + i(ad+bc)**

Thus multiplication of two complex number results in above expression.

My suggestion is to remember the formula as it will help you solve the multiplication faster. i hope you have understood the concept of multiplication of complex number, let us solve some questions for better understanding.

**Example 01**

z1 = 5 + 2i

z2 = 10 + 4i

Find z1 * z2**Solution**

Using the multiplication formula of complex number**z1 * z2 = (ac – bd) + i(ad+bc)**

From the question, you can find the values of a, b, c, d

a = 5

b= 2

c= 10

d= 4

Putting the value in the formula

z1 * z2 = (50 – 8) + i(20+20)**z1 * z2 = 42 + 40i**

**Example 02**

z1 = -4 + 2i

z2 = 2 + 8i

Find z1 * z2**Solution**

Using the multiplication formula of complex number**z1 * z2 = (ac – bd) + i(ad+bc)**

here

a = -4

b = 2

c = 2

d = 8

Putting the values

z1 *z2 =(-8-16) + i(-32+4) **z1 *z2 = -24 -28i**

**Properties of multiplication of complex number**

Here i have listed some of the properties related to multiplication of complex number. Understanding these properties will enhance your understanding of this chapter.

**(a) Closure law of multiplication**

Multiplication of complex number will also results in complex number

z1 * z2 =z3

If z1 and z2 are complex number, then the result z3 will also be a complex number

**(b) Commutative law**

Interchanging the position of complex number in the multiplication will have no impact in the final result**z1 * z2 = z2 * z1**

**(c)Associative Law**

In multiplication of three complex number, formation of different grouping in the process will have no impact on the final result**(z1 * z2) z3 = z1 (z2 * z3**)

**(d) Multiplicative inverse**

The multiplicative inverse of complex number z= a + ib can be written as:

\frac { 1 }{ z } =\frac { a }{ { a }^{ 2 }+{ b }^{ 2 } } +i\frac { -b }{ { a }^{ 2 }+{ b }^{ 2 } }

This is called multiplicative inverse because z * \frac { 1 }{ z } =1

**(e)Distributive Law**

The distributive law for any three complex numbers z1, z2, z3

**Division of complex Numbers**

Division of complex number is not as simple as addition or multiplication of complex number. It requires more calculation and you have to be careful with the steps involved.

Suppose there are two complex number z1 and z2 such that

z1 = a + ib

z2 = c + id

And you have been asked to find z1/z2, then follow the below steps**Step 1**

For calculation of z1/z2, first take multiplicative inverse of 1/z2

\frac { 1 }{ z2 } =z2\quad (multiplicative\quad inverse)\quad

**Step 02**

Then multiply z1 *z2

Let us understand the division of complex number with the help of examples

**Example 01**

z1= 6 + 3i

z2 = 2 – i

Find z1/z2

**Solution**

This question involve division of complex number so we will follow the above mentioned steps

Step 01: Calculate multiplicative inverse of 1/z2

z2= \frac { a }{ { a }^{ 2 }+{ b }^{ 2 } } +i\frac { -b }{ { a }^{ 2 }+{ b }^{ 2 } }\\\ \\
z2= \frac { 2 }{ { 2 }^{ 2 }+{ (-1) }^{ 2 } } +i\frac { -(-1) }{ { 2 }^{ 2 }+{ (-1) }^{ 2 } } \\\ \\ z2= \frac { 2+i }{ 5 }

Step 02: Now we calculate z1 * z2

\ z1\times z2=\quad (6+3i)\times \frac { 2+i }{ 5 } \\\ \\ \\\ \\ z1\times z2=\frac { 1 }{ 5 } (9+12i)\