# Find a conformal map between two sectors.

It is clear that the module of annular regions $v=\{(r,\theta): r_1<r<r_2, a<\theta<b\}$ is $m=\frac{r2}{r1}$. But how can I find a conformal mapping between two sector $A=\left(r_1<r<r_2, \frac{-\pi}{4}< \theta < \frac{\pi}{4} \right)$ and $B=\left(R_1 < r < R_2, \frac{-\pi}{2}< \theta < \frac{\pi}{2} \right)$ which have different angle ? Thanks.. • What do you mean by the "module" of a region??? Mar 5, 2018 at 15:38
• math.stackexchange.com/questions/2222319/… Mar 6, 2018 at 22:59
• Dear David, i must be find a conformal mapping rule for two sectorial regions which have different sector angles. i is very crucial for me. Mar 6, 2018 at 23:05
• @halis That link is for the module of an annulus, it does not apply to an annulus region, so your question is ill-founded. May 15, 2018 at 15:51

Hint: use $$z\longmapsto a(bz)^\alpha.$$ You can determinate $a,b,\alpha$ using polar coordinates.
EDIT: I understand that the "moduli" (your definition $\ne$ the link definition) of both regions are related as $M = m^2$.
By the geometric interpretation of the product of two complex numbers (argument of product is...), $\alpha = 2$ obviously. We want also $$|z| = r_i\implies |a||b|^2|z|^2 = |a(bz)^2| = R_i,\qquad i = 1,2.$$ As no rotation required, $a$ and $b$ will be real an positive. This implies $$ab^2 r_i^2 = R_i,\qquad i = 1,2.$$ $$ab^2 = \frac{R_1}{r_1^2} = \frac{R_2}{r_2^2}.$$