tan{i log (a−iba+ib)}=

Let a+ib=rcisθ=reiθ so that a=r cosθ,b=rsinθ where r=a2+b2,tanθ=b/aThen a−ib=rcis(−θ)=r e−iθtan[i log(a−iba+ib)]=tan{i log (e−iθeiθ)}=tan{i log e−2/θ}=tan2θ=2tanθ1−tan2θ=2aba2−b2.

tan{i log (a−iba+ib)}=

Let a+ib=rcisθ=reiθ so that a=r cosθ,b=rsinθ where r=a2+b2,tanθ=b/aThen a−ib=rcis(−θ)=r e−iθtan[i log(a−iba+ib)]=tan{i log (e−iθeiθ)}=tan{i log e−2/θ}=tan2θ=2tanθ1−tan2θ=2aba2−b2.

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$\tan {i \log (\frac{a - i b}{a + i b})} =$

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$a b$

$\frac{2 a b}{a^{2} - b^{2}}$

$\frac{a^{2} - b^{2}}{2 a b}$

$\frac{2 a b}{a^{2} + b^{2}}$

answer is B.

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Detailed Solution

Let $a + i b = r c i s θ = r e^{i θ}$ so that $a = r \cos θ, b = r \sin θ$ where $r = \sqrt{a^{2} + b^{2}}, \tan θ = b / a$

Then $a - i b = r c i s (- θ) = r e^{- i θ}$

$\tan [i \log (\frac{a - i b}{a + i b})] = \tan {i \log (\frac{e^{- i θ}}{e^{i θ}})} = \tan {i \log e^{- 2 / θ}} = \tan 2 θ = \frac{2 \tan θ}{1 - \tan^{2} θ} = \frac{2 a b}{a^{2} - b^{2}} .$