# 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.

• The image of $z-1/z$ should be such that for $w$ in the image, $\sqrt{w^2+4}$ is well defined for all points for it to be biholomorphism. How can we think of the image of our domain under $z-1/z$? Commented Jun 3 at 11:32

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.