Finding an Arg(w) The question is: 
Describe (in words) and sketch the set of all $z \in \mathbb{C}$
such that
$$\displaystyle 0<\arg\left(\frac{i-z}{i+z}\right)<\frac{\pi}{2}$$
I believe that I am supposed to break down $\displaystyle \frac{i-z}{i+z}$ first, but I am having trouble.
 A: Let 
$$
\begin{align}
w=&\frac{i-z}{i+z}=\lbrace\text{with }z=x+iy\rbrace=\frac{-x+i(1-y)}{x+i(1+y)}=\frac{(-x+i(1-y))\cdot(x-i(1+y))}{(x+i(1+y))\cdot(x-i(1+y))}
\\=&\frac{(1-x^2-y^2)+i(2x)}{x^2+(1+y)^2}=\frac{1-x^2-y^2}{x^2+(1+y)^2}+i\frac{2x}{x^2+(1+y)^2}
\end{align}$$
For $\arg(w)$ to be in the interval $(0,\pi/2)$, we have that $w$ is in the first quadrant, or equivalently: $\Re(w)>0$ and $\Im(w)>0$. Thus we get from the calculations above, that, as long as $(x,y)\neq(0,-1)\Leftrightarrow z\neq -i$, we have:
$\Re(w)=\frac{1-x^2-y^2}{x^2+(1+y)^2}>0\Leftrightarrow1-x^2-y^2>0\Leftrightarrow x^2+y^2<1$
and
$\Im(w)=\frac{2x}{x^2+(1+y)^2}>0\Leftrightarrow 2x>0\Leftrightarrow x>0$.
Now it's just a matter of sketching left to see what OP's set looks like.
A: Circles and lines are mapped to circles are lines for such transformations.
It is easy to see that $z=it$ is mapped to $(i-it)/(i+it)=(1-t)/(1+t)$. So the imaginary axis is mapped to the real axis, with $t\in[1,-1)$ mapped to the positive real axis. The unit circle must be mapped to the imaginary axis because it is mapped to a line (because it passes through $-i$) that meets the real axis orthogonally at the origin. Because of orientation, the right half of the unit circle traversed from top to bottom must be mapped to the positive imaginary axis from origin to $+i\infty$. So, it appears to me that $\{ z : \Re z > 0, \; |z| < 1 \}$ is mapped to the first quadrant where $0 < \arg < \pi/2$.
Verification: Let's check this. First, $|z| < 1$ should be mapped to the right half plane:
$$
       \Re \frac{i-z}{i+z} = \Re \frac{(i-z)(-i+\overline{z})}{|i+z|^{2}}
    = \Re \frac{(1+iz+i\overline{z}-|z|^{2})}{|i+z|^{2}}=\frac{(1-|z|^{2})}{|i+z|^{2}} 
$$
So that checks. And the imaginary part should be positive if $|z| < 1$ and $\Re z > 0$. Let's verify:
$$
      \Im \frac{i-z}{i+z} =\Im \frac{(i-z)(-i+\overline{z})}{|i+z|^{2}}
    = \Im \frac{(1+iz+i\overline{z}-|z|^{2})}{|i+z|^{2}}=\frac{2\Re z}{|i+z|^{2}} 
$$
Everything checks.
