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

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?

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.