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.
Best Answer
$R$ 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.