Mapping circular lunes $|z|<1,|z+i|<1$

Question

I want to map the circular lunes $|z|<1,|z+i|<1$ onto the upper half-plane

Since intersection points are $z_1=\sqrt 3/2-i/2$ and $z_2=-\sqrt 3/2-i/2$ we have

$$w_1=\frac {2z-\sqrt 3+i}{2z+\sqrt 3+i} $$

Now $w_1$ maps the lunes to the region between the lines $v=+\sqrt 3 u $ and $v=-\sqrt 3 u $ in left half plane, that is

$$2\pi/3 <arg w < 4\pi/3$$

Now $e^{-i2\pi/3}w_1$ maps to

$$0 <arg w < 2\pi/3$$ and rising to power $3/2$ we arrive to upper half plane.

So my answer is $e^{-i\pi}w_1^{3/2}$, but the answer in end of my book is $e^{i\pi/3}w_1^2$. Could anyone please say me if my answer is wrong or my book is wrong? Thanks!

Answer 1

Your answer is correct. Note that the inner angle of the lens at $z_1$ and $z_2$ is ${2\pi\over3}$. This angle remains after the Möbius transformation $z\mapsto w$. Only the exponent ${3\over2}$ can convert this angle to $\pi$, as required for a half plane.