Find the principal argument of a complex number I have a text book question to find the principal argument of
$$ z = {i \over -2-2i}. $$
I know formulas where we find using $$ \tan^{-1} {y \over x}$$
but I am kinda stuck here can somebody please help.
 A: Rewrite as
$$z = -\frac{1+i}{4}$$
Note that $\Re{z} = \Im{z} = -1/4$, so that the argument of $z$ lies in the third quadrant (same sign, both negative).  Because the real and imaginary parts are equal,
$$\text{Arg}\,{z} = \frac{-3 \pi}{4}$$
A: $$z=\frac i{-2-2i}=\frac12\frac {-i}{1+i}=\frac12\frac {-i(1-i)}{(1+i)(1-i)}=\left(-\frac14\right)+i\left(-\frac14\right)$$
Using the definition of $\arctan2,$
the principal value of the Argument of $z$ 
(which lies in $(-\pi,\pi]$) will be $\arctan 1 -\pi=\frac\pi4-\pi=-\frac{3\pi}4$  (as $\frac{\left(-\frac14\right)}{\left(-\frac14\right)}=1$)

Alternatively, using this, $$Arg \left(\frac {z_1z_2\cdots}{w_1w_2\cdots}\right)=\sum Arg(z_i)-\sum Arg(w_i)\pmod {(-\pi,\pi]}$$
$$\text{So, }Arg \left(\frac i{-2-2i}\right)= Arg(i)-Arg(-1)-Arg(1+i)$$
$$=\frac\pi2-\pi-\frac\pi4\pmod {(-\pi,\pi]}=-\frac{3\pi}4$$
A: $z = \cfrac{i}{ -2-2i}=\cfrac{i(-2+2i)}{ (-2-2i)(-2+2i)}=\cfrac{i(-2+2i)}{ (-2-2i)\overline{(-2-2i)}}=\cfrac{i(-2+2i)}{ \left|-2-2i\right|^2}=\cfrac{-2i-2}{ \left|-2-2i\right|^2}=\cfrac{1}{ \left|-2-2i\right|^2}(-2-2i)$
Since $\cfrac{1}{ \left|-2-2i\right|^2}\in \Bbb R_+^*$, the principal argument of $z$ is also the principal argument of $-2-2i$ which you should be able to find.
A: After you get to the point where you write $z = -\cfrac{1}{2}-\cfrac{\mathrm{i}}{2}$, you can rewrite this as $z = (\cfrac{1}{2\sqrt 2})(-\cfrac{\sqrt 2}{2}-\cfrac{\mathrm{i}\sqrt 2}{2})$. You can then use the formula $z = r\mathrm{e}^{\mathrm{i}\theta}$, where $r = \lvert z \rvert$, and $\mathrm{e}^{\mathrm{i}\theta} = \cos\theta + \mathrm{i}\sin\theta$, where $\theta = arg(z)$. Here $\theta = \cfrac{5\pi}{4}+2k\pi, k \in \Bbb Z$. If $Arg(z) \in [0, 2\pi)$, $Arg(z) = \cfrac{5\pi}{4}$. If $Arg(z) \in (-\pi, \pi]$, $Arg(z) = -\cfrac{3\pi}{4}$.
