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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$\log 2$

Vedclass · Answer

(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

Similar Questions

If $x=\sin \left(2 \tan ^{-1} 2\right)$ and $y=\sin \left(\frac{1}{2} \tan ^{-1} \frac{4}{3}\right)$,then

$x, y, z$ are in $G$.$P$. and $\tan^{-1} x, \tan^{-1} y, \tan^{-1} z$ are in $A$.$P$.,then

$\lim _{n \rightarrow \infty} \tan \left\{\sum_{r=1}^{n} \tan ^{-1}\left(\frac{1}{1+r+r^{2}}\right)\right\}$ is equal to..........

If $u = \tan^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right)$ and $v = 2 \tan^{-1} x$,then $\frac{du}{dv}$ is equal to

If $y = \tan^{-1}(\sec x - \tan x)$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module