prove the following equation about inverse of tan in logarithmic for

Question

$$\arctan(z)=\frac1{2i}\log\left(\frac{1+iz}{1-iz}\right)$$

i have tried but my answer doesn't matches to the equation .the componendo dividendo property might have been used. where $$\arcsin(x)=\frac1i\log\left(iz+\sqrt{1-z^2}\right)$$

Answer 1

Using Euler's formula:- $$e^{i\theta}=\cos\theta+i\sin\theta$$ we have $$2i\sin\theta=e^{i\theta}-e^{-i\theta}\text{ (1)}$$ $$2\cos\theta=e^{i\theta}+e^{-i\theta}\text{ (2)}$$

Dividing Equation $(1)$ by $(2)$ and then dividing both sides by $i$ results in $$\tan\theta=\frac{1}{i}\left(\frac{e^{i\theta}-e^{-i\theta}}{e^{i\theta}+e^{-i\theta}}\right)$$ Let $z=\tan\theta \Leftrightarrow \theta=\arctan z$, so that

$$z=\frac{1}{i}\left(\frac{e^{i\theta}-e^{-i\theta}}{e^{i\theta}+e^{-i\theta}}\right)\\\color{blue}{\Rightarrow iz(e^{i\theta}+e^{-i\theta})=e^{i\theta}-e^{-i\theta}}\\\Rightarrow e^{i\theta}(1-iz)=e^{-i\theta}(1+iz) \\\Rightarrow e^{2i\theta}=\frac{1+iz}{1-iz}\\\Rightarrow 2i\theta=\log\left(\frac{1+iz}{1-iz}\right)\\\Rightarrow \theta=\frac{1}{2i}\log\left(\frac{1+iz}{1-iz}\right)\\\Rightarrow \arctan z=\frac{1}{2i}\log\left(\frac{1+iz}{1-iz}\right)$$

Answer 2

Let $\arctan(z) = t$. We then have $z = \tan(t)$. Now look at the right hand side and we have, $$\dfrac1{2i} \log\left(\dfrac{1+iz}{1-iz}\right) = \dfrac1{2i} \log\left(\dfrac{1+i\tan(t)}{1-i\tan(t)}\right) = \dfrac1{2i} \log\left(\dfrac{\cos(t)+i\sin(t)}{\cos(t)-i\sin(t)}\right) = \dfrac1{2i} \log\left( \dfrac{e^{it}}{e^{-it}}\right) = t$$