The operation of complex numbers on Subtraction is simple in a way that one have to do the normal subtraction

**Example: 1.**

**(a-ib)-(c-id) . **In this example, one is required to separate between **Real part and Imaginary part.** *By doing so you’ll have (a-c)-i(b-d)*

**i.e**

**(a-ib)-(c-id) = (a-c)-i(b-d) **

**Example: 2.**

** (4-5i) – (3+2i) = (4-3) – i(5-2)**

** = 1 -3i**

**Example: 3. **

** (7+5i) – (3+2i) = (7-3) + i(5-2)**

**= 4 + 3i **

**Note: ***For one solve this problem, first put the like term together and then solve. That is, separate the Imaginary part from** Real part.*

Check out this video for more

https://www.youtube.com/watch?v=6CumaOLGDYQ

On the Hand, Operation on Division is a bit complicated and care should be taken so as to avoid unnecessary error or mistake.

Operation on Division of Complex number is the form ** (a+ib)/(c-id). **Solving this problem fully, one has to conjugate (*to multiply by the conjugate both numerator and denominator* )

**Example:**

**(a+ib)/(c+id) = {(a+ib)(c-id)}/{(c+id)(c-id)}**

**= {(ac – iad + ibc + bd)}/{c^2 + d^2}**

** = {(ac+bd)+i(bc-ad)}/ {c^2 + d^2}**

= **[{(ac+bd)}/ {c^2 + d^2} ] + [i{(bc-ad)}/ {c^2 + d^2]**

2**. (2+3i)/(4+3i) = {(2+3i)(4-3i)}/{(4+3i)(4-3i)} ***Note: the conjugate of 4+3i is 4-3i*

= **{(2+3i)(4-3i)}/{25} ***the multiplication of complex number and its conjugate gives real number*

**=(17 + 6i)/25**, *at this point you can separate the real and Imaginary part, to get*

= **(17/25) + i(6/25), ***the final answer.*

Check out this video for more information

https://www.youtube.com/watch?v=o5oE2K-t5Qw&t=22s

My name is Benson Karue, I was Software Engineer Telenet Co LTD before I quit and joined YouTube and Udemy as an instructor. I have real-world software experience and have a Bachelor’s degree in Mathematics and Computer Science with 6+ years of coding experience and winning several competitions and coding challenges.

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