What is the domain $-\pi \leq \arg(z+i) \leq \frac{\pi}{2}$? What is the domain $-\pi \leq \arg(z+i) \leq \frac \pi 2$?
a simple math shows that:
$$-\pi \leq \arctan \left(\frac{y+1} x \right) \leq \frac \pi 2$$
so that leads to: (by applying $\tan(x)$ on the whole inequality) 
$$0 \leq \frac{y+1}{x} \leq \infty$$
and I concluded that the domains are: $\{y<-1,x<0\}$, $\{-1<y<0, x>0\}$. or $\{y>0,x>0\},$ but looks like this should work also for $\{x>0\}$, and I can't see why.

Consider 
$$
-\pi\leq \operatorname{arg}(z)\leq 0 \Leftrightarrow \Im(z)\leq 0
$$
and
$$
-\frac{\pi}2\leq \operatorname{arg}(z)\leq \frac{\pi}2\Leftrightarrow \Re(z)\geq 0.
$$
Using the cartesian coordinates $z=x+yi$ you get $z+i=x+(1+y)i$ and
\begin{align}
\left\{z\in\mathbb{C}~:~-\pi\leq\operatorname{arg}(z+i)\leq\frac{\pi}2\right\}&=\left\{z\in\mathbb{C}~:~-\pi\leq\operatorname{arg}(z+i)\leq0\vee-\frac{\pi}2\leq \operatorname{arg}(z+i)\leq\frac{\pi}2\right\}\\
&=\left\{z\in\mathbb{C}~:~\Im(z+i)\leq 0~\vee~\Re(z+i)\geq 0\right\}\\
&=\left\{x+yi\in\mathbb{C}~:~y+1\leq 0~\vee~x\geq 0\right\}\\
&=\left\{x+yi\in\mathbb{C}~:~y\leq -1~\vee~x\geq 0\right\}
\end{align}
