continuous determination of the argument Let $U=\{z=|z|e^{i\theta} :\theta \in (-\pi,\pi)\}$ and  $V=\{z\in \mathbb C : -\pi< \Im(z) <\pi \}$, and $$\log_{\pi}:U \to V  : z=\rho e^{i\theta} \mapsto \ln(\rho)+i\theta$$ the branch of complex logarithm,
with $z=x+iy$ and $$ \theta=
\begin{cases}
\arctan(y/x), & \text{if } x>0\\
\pi+\arctan(y/x), & \text{if } x<0,y>0\\
-\pi+\arctan(y/x), & \text{if } x<0,y<0
\end{cases}$$
I don't know where this values of  $\theta$ comes from. I inderstand the first case where $x>0$ but the I didn't understand the other two cases ($x<0,y>0 $ and $x<0,y<0$) ! So any help is highly appreciated!
 A: This may help:
$$ \theta=
\begin{cases}
\arctan(y/x), & \text{if } (x,y)\in Q_1\cup Q_4\\
\pi+\arctan(y/x), & \text{if } (x,y)\in Q_2\\
-\pi+\arctan(y/x), & \text{if } (x,y)\in Q_3
\end{cases}$$
Since $\text{arctan}$ has range $(-\pi/2,\pi/2)$, in order to recover $\theta$ in $Q_2$ or $Q_3$ you need to shift it by $\pi$.  And although it doesn't really matter which way you shift it (plus or minus), they have chosen to do it so that $\theta$ is expressed in the interval $(-\pi,\pi)$.
A: Do you understand that, because the tangent function has period $\pi$ (rather than $2\pi$ like sine or cosine) the arctan function can only give values from $-\pi/2$ to $\pi/2$, the first and fourth quadrants.
For example, if $x= -3$ and $y= 3$, in the second quadrant, then $\arctan(y/x)= \arctan(-1)= -\pi/4$ which is in the fourth quadrant, not the second.  You have to add $\pi$ to move it to the second quadrant.  Similarly, if $(x, y)= (-3, -3)$, in the third quadrant, the $y/x= -3/-3= 1$ exactly as if it were $(3, 3)$!  The arctangent is $\pi/4$ in the first quadrant.  Adding $\pi$ moves it to the third quadrant.
