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

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 )


  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

Also read here

Subtraction and Division of Complex Numbers

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