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Question

cos(ilogaiba+ib) is equal to


A
ab
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B
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C
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D
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Solution

The correct option is B
Let  a+ib=reiθ            aib=reiθilogaiba+ib=ilog(e2θi)=i(2θi)loge=2θcos(ilogaiba+ib)=cos2θ=1tan2θ1+tan2θ=1b2a21+b2a2=a2b2a2+b2

Mathematics

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