I cannot calculate $\tan^{-1}(1+i)$ I use the formula below for inverse tangent function:
$$\tan^{-1}(z)= \frac {i} 2 \log \biggr( \frac {i+z}{i-z} \biggr)$$
I have written 
$$\tan^{-1}(1+i)= \frac {i} 2 \log \biggr( \frac {i+1+i}{i-1-i} \biggr)=  \frac {i} 2 \log (-2i-1)$$
The answer is $n\pi i , n\in \mathbb Z$ but I don’t know or I cannot see what is the argument of $-2i-1$
I know it is basic question but I couldn’t see how can I do
Thanks for helps

The number $-1-2i$ is in the third quadrant, so its argument, taken in $[0,2\pi)$, is
$$
\theta=\pi+\arctan\frac{-2}{-1}=\pi+\arctan2
$$
(it would be wrong to say that the argument is $\arctan2$). Thus
$$
-1-2i=\sqrt{5}e^{i\theta}
$$
and its logarithms are
$$
\frac{1}{2}\log5+i\theta+2k\pi i
$$
and therefore
$$
\frac{i}{2}\left(\frac{1}{2}\log5+i\theta+2k\pi i\right)=-\theta-k\pi + \frac{i}{4}\log5=
-\frac{\pi}{2}-\frac{1}{2}\arctan2-k\pi + \frac{i}{4}\log5
$$
that can also be written as
$$
\frac{\pi}{2}-\frac{1}{2}\arctan2+k\pi+\frac{i}{4}\log5
$$
