finding $\tan^{-1}(2i)$

Question

I'm having some trouble finding $\tan^{-1}(2i)$. The formula the book has is $\tan^{-1}z=\dfrac{i}{2} \log\dfrac{i+z}{i-z}$. But when I use this I get a different answer than what the book has. This is what I have so far

$\log\dfrac{i+2i}{i-2i}=\log(-3)=\ln3+i(\pi+2\pi n)$ where $n\in\mathbb{Z}$. Thus $\dfrac{i}{2} \log (-3)=\dfrac{i}{2}(\ln3+i(\pi+2\pi n))=i\dfrac{\ln 3}{2}-(\dfrac{\pi}{2}+\pi n)$. But instead of the minus sign between the two terms it is supposed to be plus which I'm not getting. I'm sure there is a mistake somewhere which i'm not able to see.

Answer 1

I will follow another method:

Let $\displaystyle\tan^{-1}(2i)=a+ib$ where $a,b$ are real

$\displaystyle\implies \tan(a+ib)=2i\iff\tan(a-ib)=-2i$

So, $\displaystyle\tan2a=\tan(a+ib+a-ib)=\frac{\tan(a+ib)+\tan(a-ib)}{1-\tan(a-ib)\tan(a+ib)}=0$

So, $2a=n\pi$ where $n$ is any integer

Similarly, $\displaystyle\tan2ib=\tan[a+ib-(a-ib)]=\frac{\tan(a+ib)-\tan(a-ib)}{1+\tan(a-ib)\tan(a+ib)}$

$\displaystyle\implies i\tanh 2b=\frac{4i}5$

Now, $\displaystyle \tanh2b=\frac{e^{2b}-e^{-2b}}{e^{2b}+e^{-2b}} \implies\frac{e^{2b}-e^{-2b}}{e^{2b}+e^{-2b}}=\frac45$

Applying componendo & Dividendo, $\displaystyle\frac{e^{2b}}{e^{-2b}}=9\iff4b=\ln9=2\ln3$

Answer 2

Note that $-(\frac {\pi}2+\pi n)=\frac {\pi}2+(-1-n)\pi$ so the two forms are equivalent