Operation of Complex Numbers on Division and Substraction

1 Comment / complex analysis / By

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

Also read here

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:

  1. (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

Also read here

Subtraction and Division of Complex Numbers



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