# Conformal map from the intersection of two disks to the unit disc

I'm working through past papers for a Complex Analysis class I'm taking and have come across the following problem on conformal maps:

Let $\lambda=\frac{1}{2}(1+i \sqrt3)$ and $R$ be the region

$$r=\{z \in \mathbb{C} : |z-\lambda|<1 \text{ and } |z-\bar{\lambda}|<1 \}$$

Determine the image of $R$ under the Möbius Transformation

$$f(z)=\frac{z}{1-z}$$

and hence find a holomorphic bijection $h$ from $R$ to the unit disc $\mathbb{D}= \{z \in \mathbb{C} : |z|<1 \}$

Thoughts

I'm not very comfortable with mapping lens-shaped regions. I think it's significant that as $1$ lies inside $R$, and this is clearly mapping to $\infty$ by $f$ in the extended complex plane, and $0$ is mapped to itself, perhaps this will be the UHP or something similar, though I'm really not sure.

Any help would be very appreciated. Best, MM.

$R$ is the intersection of two disks, and so $f(R)$ is the intersection of two images of disks. If you find those images individually, you should be well on your way to solving the problem.
• So if we call the discs $D_1$, $D_2$ then we can see $D_1$ intersects the axes at $0, 1, \sqrt 3$ and $D_2$ intersects the axes at $0, 1, - \sqrt 3$. So $0 \rightarrow 0$ and $1 \rightarrow \infty$ and $\pm \sqrt 3 \rightarrow (- \frac{3}{4} \pm i \frac{\sqrt 3}{4})$, so is $f(R)$ a 'wedge' for points with argument between $\pm \frac{5 \pi}{6}$ ? – Mathmo Jan 13 '12 at 10:06
• I get that it's the wegde for points with argument between $\pm\frac\pi6$. Maybe there's a small mistake somewhere - but the method seems sound. – Greg Martin Jan 13 '12 at 19:53