This is taken from Rudin's Real and Complex Analysis: For $0 < \alpha < 1$, define $\Omega_{a}$ to be the union of the disc $D_{\alpha}(0)$ and the line segments from $z = 1$ to points on the disc. The region so obtained is an open set bounded by the disc and the tangents from $z=1$ to it. If $z=re^{is}\in D_{\alpha}(0)$, it is easy to show that $\frac{|r-z|}{1-r}$ is bounded. Now I want to show that this is true for all $z\in \Omega_{a}$.
Wlog $z=re^{is}\in \Omega^+_{a}$, and not in the disc. Let $w=|w|e^{is}$ be the point lying on the tangent line from $z=1$ to the disc. Then, $\Re(r-z)< \Re w- \Re z=(|w|-r)\cos s$ and $\Im (r-z)< \Im w.$ Also, $1-r>1- \Re w=1-|w|\cos s$ so a calculation gives
$\tag 1\left ( \frac{|r-z|}{1-r} \right )^{2}\le \left ( \frac{\Re z-|w|}{1- \Re w} \right )^2.$
I found the equation of the tangent line in terms of $\alpha,$ using the fact that if $\theta $ is the angle whose vertex is $z=1,$ between the horizontal axis and the tangent line, then $\sin \theta =\alpha$ and after several substitutions into $(1)$ I got a mess. I thought this exercise would be easy, but I am sorry to say, I am stuck.
Edit: if $t$ is the arc subtended by $s$, and if $y$ is the length of the perpendicular drawn to $z=r$ from the line segment connecting $z=1$ to the disc, then
$s=\frac{t}{r}\le \frac{y}{r}\ $ and $\ y=(1-r)\tan \theta=(1-r)\cdot \frac{\alpha }{\sqrt{1-\alpha ^{2}}}.$
Using these facts, and recalling that $r>\alpha,$ we can now find a bound which depends only on $\alpha:$
$\frac{|z-r|}{1-r}=\frac{r}{1-r}\cdot |1-e^{-is}|\le \frac{r}{1-r}\cdot \sum_{k=1}^{\infty}\frac{s^{k}}{k!}\le \frac{r}{1-r}\cdot \sum_{k=1}^{\infty}\frac{\left (\left ( \frac{1-r}{r} \right ) \cdot \frac{\alpha }{\sqrt{1-\alpha ^{2}}} \right )^k}{k!}=$
$\frac{\alpha }{\sqrt{1-\alpha ^{2}}}+\sum_{k=2}^{\infty}\frac{(\frac{1-r}{r})^{k-1}\cdot (\frac{\alpha }{\sqrt{1-\alpha ^{2}}})^{k}}{k!}$.