Find a Mobius transformation sending the region $D$ between $|z-1|=1$ and $|z|=1$ to ${0<Arg(z)<\frac{2\pi}{3}}$ such that 1 is mapped to $i$.
My idea: The two circles intersect at $x+iy= \frac{1}{2} \pm \frac{\sqrt{3}}{2}i$, while the two rays of the wedge intersect at $0$ and $\infty$. I therefore thought that a reasonable way to approach this is to find a Mobius map with the following properties: $\frac{1}{2} - \frac{\sqrt{3}}{2}i \to 0$ , $1\to i$, $\frac{1}{2} + \frac{\sqrt{3}}{2}i\to \infty$ and then find the map using the cross-ratio preservation: $$ \frac{z-1}{\left( \frac{1}{2} - \frac{\sqrt{3}}{2}i \right)-1} \frac{\left( \frac{1}{2} - \frac{\sqrt{3}}{2}i \right)-\left( \frac{1}{2} + \frac{\sqrt{3}}{2}i \right)}{z-\left( \frac{1}{2} - \frac{\sqrt{3}}{2}i \right)-\left( \frac{1}{2} + \frac{\sqrt{3}}{2}i \right)} = \frac{M(z)-i}{0-i}, $$ where $M$ is the Mobius transformation I am looking for.
My question is:
What is the image of the unit disc $|z|=1$ under this transformation? From a known theorem, and from $\frac{1}{2}+\frac{\sqrt{3}}{2}\to\infty$ the image should be a straight line, which seems to be the imaginary axis $Re(z)=0$. But, shouldn't the image of the unit circle be one of the two rays that form the boundary of ${0<Arg(z)<\frac{2\pi}{3}}$? I am specifically confused about the image of each part of the boundary of the domain $D$.
Thank you
Let $C_1$ be the circle $|z|=1$, $C_2$ the circle $|z-1|=1$, and denote their intersections with $a = 1/2 + i \sqrt 3/2$ and $\bar a = 1/2 - i \sqrt 3/2$.
Let us ignore the condition $M(1) = i$ for a moment, and map $D$ to a sector domain. $a$ and $\bar a$ must be mappend to $0$ and $\infty$ (in some order), so a good start is $$ T(z) = \frac{z-a}{z-\bar a} \, . $$ Then $C_1$ is mapped to a line $L_1$ through the origin and $$ T(1) = \frac{1-a}{1-\bar a} = -a = -\frac 12 - i\frac{\sqrt 3}{2}\, , $$ and $C_2$ is mapped to a line $L_2$ through the origin and $$ T(0) = \frac{a}{\bar a} = - \bar a = -\frac 12 + i\frac{\sqrt 3}{2}\, . $$ Möbius transformations are continuous and bijective functions (aka“ homeomorphisms” aka “topological isomorphisms”) from the extended complex plane onto itself. Therefore the domain $D$ is mapped to one of the four sectors delimited by the lines $L_1$ and $L_2$.
Since $T(1/2) = -1$, $D$ is mapped to the sector $$ \frac{2\pi}{3} < \arg z < \frac{4\pi}{3} \, , $$ so that $$ M(z) = e^{-2\pi i/3} T(z) $$ is a Möbius transformation which maps $D$ onto the sector $0 < \arg z < 2 \pi /3$.
Finally: $1$ is on the boundary of $D$, so that $M(1)$ is on the boundary of $M(D)$, i.e. the condition $M(1) = i$ can not be satisfied.