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?
 A: Suppose that $X$ is a star domain and let $z=(x,y)$ be the center of the star.  
Claim: For $\epsilon>0$ sufficiently small, the line between $z$ and $(\epsilon,1+\epsilon)$ or the line between $z$ and $(\epsilon,-1-\epsilon)$ intersects the unit disk.
Sketch: If the imaginary part of $z$ is greater than $0$, i.e., $y>0$, then consider the line between $(x,y)$ and $(\epsilon,-1-\epsilon)$.
Recall from calculus, the distance between a line and a point.  If the line is given by $ax+by+c=0$ and the point is $(x_0,y_0)$, then the distance is 
$$
\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}.
$$
In our case, we're interested in the distance between the line and the origin.  Therefore, $(x_0,y_0)=(0,0)$, simplifying the formula.
A vector in the direction of the line is $(x-\epsilon,y+1+\epsilon)$, so an equation for the line is
$$
(y+1+\epsilon)X+(\epsilon-x)Y+c=0.
$$
Substituting in the point $(X,Y)=(\epsilon,-1-\epsilon)$, this implies that 
$$
(y+1+\epsilon)(\epsilon)+(\epsilon-x)(-1-\epsilon)+c=0.
$$
Therefore,
\begin{align}
a&=y+1+\epsilon\\
b&=\epsilon-x\\
c&=-y\epsilon-x-x\epsilon.
\end{align}
Now, what you want to do is to show $c^2\leq a^2+b^2$ for $\epsilon$ sufficiently small.  Left to the reader.
Small caveat: You need that the imaginary part of $z$ is greater than zero in order to get that the triangle $(0,0)$, $(\epsilon,-1-\epsilon)$, and $(x,y)$ is obtuse with the large angle at the origin.  This implies that the altitude of the triangle is contained within the triangle, so the closest point on the line to the origin is on the side of the triangle between $(x,y)$ and $(\epsilon,-1-\epsilon)$.
A: 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. ..
