complex logarithms Using complex logarithms, how would I solve this 
$$\left.\frac12i\;\text{Log}\frac{1-i(1+e^{it})}{1+i(1+e^{it})}\right|_0^{2\pi}$$
would it equal;
$$ \frac12i[ ln (\sqrt2) + I arg \frac{1-2i}{1+2i} -ln (\sqrt2) - I arg \frac{1-2i}{1+2i} + 2 \pi ik]  = - \pi k$$
 A: We used inverse trigonometric function:(http://en.wikipedia.org/wiki/Inverse_trigonometric_functions)
$$\tan^{-1} \theta=\frac{i}{2}\log\frac{1-i\theta}{1+i\theta}$$
Whence $$\to \left.\frac{i}{2}\log\frac{1-i\left(1+e^{it}\right)}{1+i\left(1+e^{it}\right)}\right|_{0}^{2\pi}=\left.\tan^{-1}\left(1+e^{it}\right)\right|_{0}^{2\pi}=\tan^{-1}\left(1+e^{2\pi i}\right)-\tan^{-1}2$$
We have $$\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy}$$
Hence $$ \tan^{-1}\left(1+e^{2\pi i}\right)-\tan^{-1}(2)=\tan^{-1}\frac{e^{2\pi i}-1}{1+2e^{2\pi i}}=\tan^{-1}\frac{1-1}{1+2e^{2\pi i}}=\tan^{-1}(0)=k\pi$$
Where $k\in \mathbb{Z}$
A: I assume, you evaluate this at values $t=2\pi$ and $t=0$. Then you see immediately, that $e^{2\pi i}=e^0=1$ which means your expression vanishes.
To evaluate the complex logarithm, the simplest way is to make the argument a complex number and then use $\log z = \log |z| + i \arg z + 2ik\pi$, $k\in \mathrm{Z}$. Here 
$z=\frac{1-2i}{1+2i}=-\frac{3+4i}{5} \ \Rightarrow\ |z|=1 \ ,\ \arg z =\tan \frac{4}{3} \ . $
A: Right. The function under the logarithm is periodic and as your bounds cover a period, the difference is just the indeterminacy on the argument of the logarithm, $2ik\pi$, which scaled by $i/2$ and ignoring the unessential minus gives 
$$k\pi.$$
