<s>[INST] Finding values of $\tan^{-1} (2i)$. I am trying to find all solutions of $\tan^{-1} (2i)$. I don't see anything that I have done wrong, my answer doesn't match the one in the textbook. Here is what I have. (The convention in Brown and Churchill is to use $\log$ for a complex number and $\ln$ for a real number.) 
\begin{align*}
\tan^{-1} (2i) & = \frac{i}{2} \log \frac{i + 2i}{i - 2i} \\
& = \frac{i}{2} \log \frac{3i}{-i} \\
& = \frac{i}{2} \log (-3) \\
& = \frac{i}{2} \left(\ln 3 + (2n + 1) \pi i    \right) \\
& = \frac{i}{2} \ln 3 - \frac{2n + 1}{2} \pi \\
& = \frac{i}{2} \ln 3 - \left(n + \frac{1}{2}\right) \pi.
\end{align*} 
The answer in the textbook, however, is: 
\begin{align*}
\frac{i}{2} \ln 3 + \left(n + \frac{1}{2}\right)\pi.
\end{align*}
This leads me to believe that I have misplaced a sign somewhere, but I cannot see where. Might there just be a typo in the book? 
Thanks in advance. 
 [/INST] Since $\tan(x)=2i$
Using Euler's formula
$\tan(x)={\frac {e^{ix}-e^{-ix}}{i(e^{ix}+e^{-ix})}}$  witch must equal $2i$
Manipulating a bit we get $3e^{ix}+e^{-ix}=0$ and if $e^{ix}=y$, $3y+\frac{1}{y}=0$ thus $e^{ix}=y=±i \frac{\sqrt 3}{3}$ 
This, of course, means that $\ln\left(±i \frac{\sqrt 3}{3}\right)=ix$
</s>