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Question

The expression tan(ilog(aiba+ib)) reduces to:



Your Answer
A

Correct Answer
B

Your Answer
C

Your Answer
D


Solution

The correct option is B


Let a+bi=reiθ. Then tanθ=ba
tan(ilogaiba+ib=tan(ilog e2iθ)
=tan(2i2θ)=2tanθ1tanθ=2ba1(b2a2)=2aba2b2

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