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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..

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halis
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1 Answers1

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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 $$ab^2 r_i^2 = R_i,\qquad i = 1,2.$$ $$ab^2 = \frac{R_1}{r_1^2} = \frac{R_2}{r_2^2}.$$

Martín-Blas Pérez Pinilla
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  • Dear Martin, can you describe it in more detail please. Because I do not know the complex analysis in detail. – halis Mar 06 '18 at 22:58
  • Thanks, but This is not depend on the sector angle. i think that it must be related with the angle because we have two sector which have two different angle. – halis Mar 12 '18 at 06:17