How to calculate complex logarithm over non usual branches I need to calculate $log$ (3 -3$i$) over the branches [0,$2$ $\pi$) , [-$\pi$, $\pi$) and [-$\pi$/2 ,3$\pi$/2)
I have no trouble  for [0,$2$ $\pi$) , [-$\pi$, $\pi$) since we have that |3-3i|= $\sqrt{18}$ , so for [-$\pi$, $\pi$)  arctg (-3/3) =$\pi$/4 then log(3+3i) =$\sqrt{18}$ + i $\pi$/4 
Also for [0,$2$ $\pi$)  I know that the arg(3+3i)=7$\pi$/4 since the argument in this branch is the angle form the origin to the position of $z$ in counterclockwise motion.
But for [-$\pi$/2 ,3$\pi$/2) I dont really now what to do , I used to think that for get the argument of $z$ in this branch we need only to add 3$\pi$/2 to the argument of this number in the principal branch  but I know this is wrong since we can only need add multiples of $2$ $\pi$ , still I dont get it
Can anyone help me to calculate the logaritm  of this number in this branch  and also explain me the background of calculating complex logarithm over non usual branches like this. Thanks

This is not hard. To compute $\log z$, just write $z$ as $re^{i\theta}$ where $\theta$ is chosen appropriately for the branch you want; then the value is $\log z = \ln r + i\theta$, where "$\ln$" is the ordinary real-valued logarithm of a positive argument. 
The "branch" corresponds to a half-open interval of length $2\pi$ so there is a unique $\theta$ in this interval for every nonzero complex number $z$.
