The imaginary part of tan^{-1}left(frac{5i}{3 | vedclass

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(C) We use the formula $	an^{-1}(iz) = i 	anh^{-1}(z) = \frac{i}{2} \log\left(\frac{1+z}{1-z}ight)$.Here,$z = \frac{5}{3}$.$	an^{-1}\left(\frac{5

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$\log 2$

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(C) We use the formula $	an^{-1}(iz) = i 	anh^{-1}(z) = \frac{i}{2} \log\left(\frac{1+z}{1-z}ight)$.Here,$z = \frac{5}{3}$.$	an^{-1}\left(\frac{5i}{3}ight) = i 	anh^{-1}\left(\frac{5}{3}ight) = \frac{i}{2} \log\left(\frac{1 + 5/3}{1 - 5/3}ight)$.$= \frac{i}{2} \log\left(\frac{8/3}{-2/3}ight) = \frac{i}{2} \log(-4)$.Since $\log(-4) = \log(4) + i(\pi + 2k\pi)$,the expression becomes $\frac{i}{2} (\log 4 + i\pi) = \frac{i}{2} \log 4 - \frac{\pi}{2}$.However,using the standard principal value branch for $	anh^{-1}(z) = \frac{1}{2} \log\left(\frac{1+z}{1-z}ight)$,we have $	an^{-1}(iz) = \frac{i}{2} \log\left(\frac{1+z}{1-z}ight)$.For $z = 5/3$,$	an^{-1}(5i/3) = \frac{i}{2} \log\left(\frac{8/3}{-2/3}ight) = \frac{i}{2} \log(-4)$.Taking the principal value,$\log(-4) = \ln 4 + i\pi$.Thus,$	an^{-1}(5i/3) = \frac{i}{2} \ln 4 - \frac{\pi}{2}$.The imaginary part is $\frac{1}{2} \ln 4 = \ln 2$.

The imaginary part of $\tan^{-1}\left(\frac{5i}{3}\right)$ is

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