Find $\tan^{-1} (i\sqrt{2})$. Problem: Find $\tan^{-1} (i\sqrt{2})$.
My attempt: We must find $z \in \mathbb{C}$ such that $\tan z = i\sqrt{2}$.
$$\tan z = \frac{\sin z}{\cos z} = i\frac{e^{-iz} - e^{iz}}{e^{iz} + e^{-iz}}$$
Let $u = e^{iz}$. Then $$i\frac{u^{-1}-u}{u+u^{-1}} = i\sqrt{2}$$ so $$1-u^2 = \sqrt{2}(u^2 + 1)$$
and
$$u^2 = \frac{1-\sqrt{2}}{1+\sqrt{2}}$$
Then we have 
$$e^{iz} = \sqrt{\frac{1-\sqrt{2}}{1+\sqrt{2}}}$$
Taking the natural logarithm of both sides gives
$$z = \frac{1}{2i}\ln\frac{1-\sqrt{2}}{1+\sqrt{2}}$$
Since $\ln$ is multi-valued and $\frac{1-\sqrt{2}}{1+\sqrt{2}}+0i=\frac{1-\sqrt{2}}{1+\sqrt{2}}e^{i2n\pi},$
we have 
$$\frac{1}{2i}\ln{\frac{1-\sqrt{2}}{1+\sqrt{2}}}=\frac{-i}{2}\texttt{Ln}\frac{1-\sqrt{2}}{1+\sqrt{2}}+ n\pi$$
That's how far I got. The book lists the final answer as $$\pi/2 + n\pi -i\texttt{Ln}(\sqrt{2}-1)$$
 A: Let $u=e^{iz}$ just as you did.
$$\begin{array}{rcl}
u^2 &=& \dfrac{1-\sqrt{2}}{1+\sqrt{2}} \\
u^2 &=& \dfrac{(1-\sqrt{2})^2}{(1+\sqrt{2})(1-\sqrt{2})} \\
u^2 &=& \dfrac{(1-\sqrt{2})^2}{-1} \\
e^{2iz} &=& -(1-\sqrt{2})^2 \\
e^{2iz} &=& (1-\sqrt{2})^2 e^{i\pi} \\
2iz &=& \ln[(1-\sqrt{2})^2e^{i\pi}] + 2ni\pi \\
2iz &=& 2\ln[\sqrt{2}-1] + i\pi + 2ni\pi \\
z &=& -i\ln[\sqrt{2}-1] + \dfrac\pi2 + n\pi \\
\end{array}$$
A: A much easier solution is as follows:
$$
\tan^{-1}(z)=-i\tanh^{-1}(iz)\\
\tanh^{-1}x=\frac{1}{2}\ln\frac{1+x}{1-x}\\
\begin{align}
\tan^{-1}(i\sqrt{2})
&=-i\tanh^{-1}(-\sqrt{2})=i\tanh^{-1}(\sqrt{2})\\
&=\frac{i}{2}[\ln(1+\sqrt{2})-\ln(1-\sqrt{2})]\\
&=\frac{i}{2}[\ln(\sqrt{2}+1)-\ln(\sqrt{2}-1)-i(\pi+2n\pi)]\\
&=\frac{\pi}{2}+n\pi+\frac{i}{2}\ln\frac{(\sqrt{2}+1)}{(\sqrt{2}-1)}\\
&=\frac{\pi}{2}+n\pi-i\ln(\sqrt{2}-1)
\end{align}
$$
since 
$$
\ln(1-\sqrt{2})=\ln[(\sqrt{2}-1)e^{i(\pi+2n\pi)}]\\
\ln\frac{(\sqrt{2}+1)}{(\sqrt{2}-1)}=-\ln[(\sqrt{2}-1)^2]=-2\ln(\sqrt{2}-1)
$$
EDIT: here is another interpretation of the result that gives a slighly different result. [This result agrees with Matlab if you type >atan(i*sqrt(2)).]
$$
\begin{align}
\tan(i\sqrt{2})
&=\frac{i}{2}\ln\frac{1+\sqrt{2}}{1-\sqrt{2}}\\
&=\frac{i}{2}\ln\left(\frac{1+\sqrt{2}}{1-\sqrt{2}}\frac{1+\sqrt{2}}{1-\sqrt{2}}\right)\\
&=\frac{i}{2}(\ln(-1)-2\ln(\sqrt{2}-1))\\
&=-\frac{\pi}{2}-n\pi-i\ln(\sqrt{2}-1)
\end{align}
$$
I just thought this was curious and you can imagine my angst when the first result disagreed with Matlab.
