Find a conformal mapping from the complex plane minus the closed unit disk, the segment $[-2,-1]$ and $[1,\infty)$ onto the unit disc

Question

I'm doing a course on Complex Analysis and as a bonus question to our set of exercises we've been asked to find a bijective conformal mapping from the complex plane minus the closed unit disk, the segment $[-2,-1]$ and $[1,\infty)$ onto the unit disc.

I'm a little stuck on the problem, my idea so far:

Use $f(z)=\frac1z$ to map my domain into the open unit disc (minus $[-1, -0.5]$, $(0,1]$) but I'm not quite sure how to deal with the 'missing' parts of my domain.

Would anyone be able to help shed some light?

Answer 1

I'd first turn the (missing) unit disk into a slit per $z\mapsto z-\frac 1z$. That leaves you with $\Bbb C\setminus [-2\tfrac12,\infty)$. Then apply $z\mapsto z+2\tfrac12$ to arrive at $\Bbb C\setminus [0,\infty)$. Then $z\mapsto -z$ to arrive at $\Bbb C\setminus(-\infty,0]$. Then $z\mapsto \sqrt z$ to arrive at the right half plane. From there, you should know the way.