Star domain of the set 
$\mathbb{C} $ \ $ \{z=x+iy \in \mathbb{C} | \, {\bf Re}\,z=0 \text{ and } {\bf Im}>0\}$
Is this a star domain and if so what is it centered at?

Any help appreciated
 A: Let's call $A=\{x+iy\in \mathbb C : Re(z)=0, Im(z)>0\}$. This set consists of the points $\{iy: y>0\}$. It is always useful to draw it.
So your set is basically $\mathbb C\setminus A$, which is all the plane except that half-line (the origin is not in $A$):

You can see that, intuitively, it is centered at the origin. Now, what's left is: pick a point $(x_0,y_0)\in \mathbb C\setminus A$, find the equation of the segment that goes through $(0,0)$ and $(x_0, y_0)$ and prove there can't be points of $A$ in it.

Pick $z_0=(x_0, y_0)\in D=\mathbb C\setminus A$. That means that $z_0$ cannot satisfy $x_0=0$ and $y_0>0$ at the same time. The segment that goes from $(x_0, y_0)$ to $(0,0)$ is the set $$[0,z_0]=\{(tx_0, ty_0), t\in[0,1]\}$$
Now, suppose that for a $t\in[0,1]$ the point $(tx_0, ty_0)$ lies in $A$. Then it must verify the following conditions:
$$(i) \ tx_0=0 \Rightarrow t=0 \text{ or } x_0=0$$
$$(ii) \ ty_0>0 \Rightarrow y_0>0$$
but if $t=0$, then the point is $(0,0)$ which we know is not in $A$. So if a point of the segment lies in $A$, then $z_0$ lies in $A$. But we picked $z_0$ in $D$, so we've arrived to a contradiction. So $[0,z_0]$ must lie in $D$.
