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.

Answer 2

Here is another possibility 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 disk.

We use the transformation $ \frac{1}{z} $ that you wanted. This maps your original region to the interior of the unit disk minus the segments $[-1, -1/2]$ and $[0, 1]$ as you remarked.

We now use our good old friend $\frac{iz + i}{-z + 1} $ to map the region to the upper half-plane. This maps our region to: $ \mathbb{H} \setminus \left( i[0, 1/3] \cup i[1, \infty] \right)$ We rotate the plane by $ e^{-i\pi/2} $ to transform it to another half-plane minus the segments: $ \left( [0, 1/3] \cup [1, \infty] \right) $. Squaring yields $ \mathbb{C} \setminus \left( (-\infty, 1/9] \cup [1, \infty) \right) $.

We are almost done if we remember how $\sin(z)$ works. In my edition of Stein's complex analysis there is a nice composition which yields $\sin(z)$ on the section named "The Dirichlet problem in a strip ". In any case the following trasnformations are motivated by having seen this in Stein.

If we displace and dilate the region by using the transformation $\frac{9}{4}(z - 5/9) $: $ \mathbb{C} \setminus \left( (-\infty, -1] \cup [1, \infty) \right) $. We now use Stein's trick and apply the $\arcsin(z)$ function, which maps the region to the infinite strip: $ \{ -\pi/2 < \Re(z) < \pi/2 \} $. Now we are almost done, we shift and rotate the strip using $ e^{i\pi/2}z + \pi i/2 $ to get: $ \{ 0 < \Im(z) < \pi \} $. Take the exponential of this region to map it to the upper half-plane. Finally, use once more our good old friend $ \frac{z - i}{z + i} $ to map the upper half-plane to the unit disk.