A) \[ab\]
B) \[\frac{2ab}{{{a}^{2}}-{{b}^{2}}}\]
C) \[\frac{{{a}^{2}}-{{b}^{2}}}{2ab}\]
D) \[\frac{2ab}{{{a}^{2}}+{{b}^{2}}}\]
E) \[{{a}^{2}}+{{b}^{2}}\]
Correct Answer: B
Solution :
\[\because \] \[\log \left( \frac{a-ib}{a+ib} \right)=\log (a-ib)-\log (a+ib)\] \[=-2i{{\tan }^{-1}}\left( \frac{b}{a} \right)\] \[\therefore \] \[-i\log \left( \frac{a-ib}{a+ib} \right)=2{{\tan }^{-1}}\left( \frac{b}{a} \right)\] \[={{\tan }^{-1}}\left( \frac{2ab}{{{a}^{2}}-{{b}^{2}}} \right)\] \[\Rightarrow \] \[\tan \left[ i\log \left( \frac{a-ib}{a+ib} \right) \right]=\frac{2ab}{{{a}^{2}}-{{b}^{2}}}\]
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