# Complex analysis - finding set and plotting

I would kindly like to ask you, if you can provide me some steps how to get this done.

Find and plot set $M =\{z \in \mathbb C:0<\arg\frac{z-i}{z+i} < \frac{\pi}{4}\}$

Thank you for your responses. (Not for sure if it is typed good because I don't know the terminology in English)

-
You've tried letting $z=x+iy$ and then trying to simplify from that? – J. M. Jan 8 '12 at 11:00
I would like to see the steps how to solve it. I am not so good at complex analysis... – velocityIs Jan 8 '12 at 11:12

This is geometry. If you draw the complex plane and call $A,B,M$ the points of affix $i,-i,z$, Then $arg \frac{z-i}{z+i}$ is the angle $(\vec{BM},\vec{AM})$, which is also the angle $(\vec{MB},\vec{MA})$.
Now, the points $M$ such that $(\vec{MB},\vec{MA}) = 0 \mod \pi$ are all on the line $(AB)$, while the points $M$ such that $(\vec{MB},\vec{MA}) = \theta \mod \pi$ are all on the circle passing through $A$ and $B$ whose center $O_\theta$ satisfies $(\vec{O_\theta B},\vec{O_\theta A}) = 2 \theta$. In both cases, crossing through A or B changes the angle by $\pi$ so you're interested in only one half of those sets.
In your case you're looking at the region delimited by the points of the line $(AB)$ that are not in the segment $[AB]$, and the points of the circle passing though $A$ and $B$ with center $O_{\pi/2}$ (whose affix is simply $-1$) who are to the left of $(AB)$.