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
