Show right half-plane with points of closed unit disk removed is not a star domain Consider the right half-plane $\{z\in {\mathbb  {C}}:{\mbox{Re}}(z)>0\}$. We define a set $X$ by removing from the right half-plane the points of the closed unit disk $\mathbb{D} = \{z \in \mathbb{C} : |z| \leq 1 \}$.
I want to show that $X$ is not a star domain, i.e. there is no point $z_0 \in X$, such that for all $z \in X$ the line segment $[z_0, z]$ is in $X$. Here's a region plot of the situation:

Intuitively the only possible choices for a $z_0$ all lie on the dashed line. However, if we were to select such a point $z^{*}$ on this line there will always be a region around the points $(0,1)$, $(0,-1)$ whose points are not joined by a line segment with $z^{*}$. In some way I think this is because the "tangents" at these points are parallel to the real axis. Can anyone help me to formalize this argument?

Intuitively, you would need to move all the way to $\infty$ on the dashed line, which isn't an option. ..
More formally, if you take a point  $z^*$ on the dashed line, the segments connecting it to $(0,1) $ and $(0,-1) $ are not tangent to the circle, for,  as you noted, the tangents are horizontal (slope $=0$).  Since the slopes are nonzero  ($m=-\frac1k$ and $m=\frac1k $), where $z^*=(k,0) $,  the segments intersect the interior of the disk...
I realize this is still kind of "hand waving"...  To be completely formal, let's use the line connecting $z^*=(k,0)$ and $(0,1) $:$$y=-\frac 1k x +1$$.
We'll show  $z=(x,y) $ on this line has $\vert z \vert = {(x^2+y^2)}^{\frac12} \lt 1$.
On the line we have, $z=(x,-\frac1k x+1$, so $\vert z \vert = \sqrt {x^2+(-\frac1k x+1)^2}=\sqrt{x^2+\frac {x^2}{k^2}-\frac2kx+1} $.  
We can square away the radical. ..  So, $$\begin {align} (1+\frac1{k^2} )x^2-\frac2kx+1\lt 1 \iff (1+\frac1{k^2})x^2-\frac2kx \lt 0 \iff (1+\frac1 {k^2})x-\frac2k \lt 0 \iff (1+\frac1 {k^2})x \lt \frac2k \iff x\lt (\frac {k^2+1}{k^2})\frac2k \iff x\lt \frac {2k^2+2}{k^3} \end {align}$$.  The RHS is positive,  since  $k \ge 1$, so just choose $x $ small enough. ..
Correction  Looks like I did it for the open disk, instead of the closed disk removed. ..   However,  I've done some searching and apparently the closure of a star domain is again a star domain . ..  This gives us the desired result, since I have (essentially ) shown the closure of the desired space isn't a star domain. ..
