SOLUTION: Imagnary numbers: {{{(1-3i)/(2+i)}}}

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Question 20910: Imagnary numbers:
%281-3i%29%2F%282%2Bi%29
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
(1-3i)/(2+i)=...we rationalise the denominator by multiplying with its conjugate...conjugate of 2+i is 2-i...hence multiplying n.r and d.r with 2-i,
(1-3i)(2-i)/(2+i)(2-i)=[1*2-1*i-3i*2+(-3i)(-i)]/[2^2-i^2]=[2-7i-3]/(4+1)
=(-1-7i)/5=-(1+7i)/5....using (a+b)(a-b)=a^2-b^2 and i^2=-1