argument and atan/arctan As I have to works with sine voltages and currents, I often have to use complex numbers. 
I know that $\tan\arg(Z)=\left(\frac{\Re(Z)}{\Im(z)}\right)^{-1}$
but, how to prove that we need to add $ \pi$ in case $\Re(Z)\leq0$ ?
I know that $\arctan(x) + \arctan(\frac{1}{x})=\pm \frac\pi2, \forall x \in \mathbb{R}^*$ but I cant go any further.
 A: That is because, when you want to determine the argument of a complex number $z=x+iy$,  really you have to solve, not a single trigonometric equation, but  a system of trigonometric equations:
$$\begin{cases}\cos\theta=\dfrac x{\sqrt{x^2+y^2}}\\[1ex] \sin\theta=\dfrac y{\sqrt{x^2+y^2}}\end{cases} $$
This system implies the equation $\;\tan\theta=\dfrac yx$, but the converse is not true, and the latter equation determines $\theta$ modulo $\pi$, whereas the system determines it modulo $2\pi$. 
To know if you have to add (or subtract a $\pi$), you can use the signs of $y$ and $y$:


*

*if $\dfrac y x >0$, either $x$ and $y$ are both positive, or they're both negative.

*if $\dfrac y x <0$, a single one of them is positive.


Thus considering the signs of $x$ and $y$ lets you know whether you  have to add $\pi$ to the solution found from  the tangent, or not.
A: That depends solely by the definition for the inverse function  $\arctan$ which is defined as the inverse of $\tan$ function on the interval $(-\pi/2,\pi/2)$. 
Therefore the derivation 
$$\theta=\tan(x/y)\implies x/y=\arctan \theta$$
is valid only for $\theta\in (-\pi/2,\pi/2)$ otherwise we need to correct by $\pi$.
A: Take, for instance, $Z=-1+i$. In this case, $\arg(Z)=\frac34\pi$. However, $\frac{\operatorname{Im}Z}{\operatorname{Re}Z}=-1$ and $\tan\frac{-\pi}4=-1$. So, both $\frac34\pi$ and $\frac{-\pi}4$ are such that its tangent is equal to $\frac{\operatorname{Im}Z}{\operatorname{Re}Z}$, but $\arg(Z)=\frac34\pi=\pi+\frac{-\pi}4$.
