A) \[\frac{2ab}{{{a}^{2}}+{{b}^{2}}}\]
B) \[\frac{{{a}^{2}}-{{b}^{2}}}{2ab}\]
C) \[\frac{2ab}{{{a}^{2}}-{{b}^{2}}}\]
D) \[ab\]
Correct Answer: C
Solution :
[c] \[\because \log \left( \frac{a-ib}{a+ib} \right)\] \[=log\left( a-ib \right)-log\left( a+ib \right)\] \[=\log ({{a}^{2}}+{{b}^{2}})-2i{{\tan }^{-1}}\left( \frac{b}{a} \right)-\left( \log {{a}^{2}}+{{b}^{2}}+i{{\tan }^{-1}}\left( \frac{b}{a} \right) \right)\]\[=-2i{{\tan }^{-1}}\left( \frac{b}{a} \right)\] \[\left[ \because \log (x+iy)=\log ({{x}^{2}}+{{y}^{2}})+i{{\tan }^{-1}}\left( \frac{y}{x} \right) \right]\] Now, \[=\tan \left\{ i\log \left( \frac{a-ib}{a+ib} \right) \right\}=\tan -\left\{ 2{{\tan }^{-1}}\left( \frac{b}{a} \right) \right\}\] \[=\tan \left\{ {{\tan }^{-1}}\left( \frac{2\frac{b}{a}}{1-\frac{{{b}^{2}}}{{{a}^{2}}}} \right) \right\}=\frac{2\frac{b}{a}}{\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}}}=\frac{2ab}{{{a}^{2}}-{{b}^{2}}}\]
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