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?

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 $(\epsilon,-1-\epsilon)$.
