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the expression tan[ilog{a-ib/a+ib)] reduces to : ans-2ab/a^2-b^2 how explain
MODULUS OF A-iB/A+iB IS 1.
THUS A-iB/A+iB CAN BE WRITTEN AS eiΘ where Θ is its argument.
on solving we get Θ = tan-(2ab/a2-b2)
thus,
tan[ilog{a-ib/a+ib)]=tan(i*log(eiΘ))=tan(i*iΘ)=tan(-Θ)=-tanΘ=2ab/a2-b2
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