A) 0
B) \[\infty \]
C) \[\log 2\]
D) \[\log 4\]
Correct Answer: C
Solution :
\[{{\tan }^{-1}}\left( \frac{5i}{3} \right)=i{{\tan }^{-1}}\left( \frac{5}{3} \right)=\frac{i}{2}\log \left( \frac{\frac{5}{3}+1}{\frac{5}{3}-1} \right)\] \[\operatorname{Im}\left( {{\tan }^{-1}}\left( \frac{5i}{3} \right) \right)=\frac{1}{2}\log 4=\frac{1}{2}.\,2\log 2=\log 2\].
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