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Yesterday I spent some time trying to find a conformal mapping which maps the region between |z + 3| < √10 and |z − 2| < √5 onto the interior of the first quadrant.

I started with determining the two intersection points of the two boundary circles, and found that they are i and - i. Next I constructed the Möbius transformation (z - i) / (z+i) which, I thought, would map the two boundary circles to two straight lines through the origin.

Furthermore I calculated the images of -3 + sqr(10) and 2 - sqr(5). The results were some real multiples of -2 + i and -3 - i. The lines through the origin and those two points didn't seem perpendicular, but the angle between them was hard to determine. Therefore I didn't know how to modify my Möbius transformation inte a conformal mapping which would satisfy the specified requirement.

My way out of the dilemma was to cheat and look at a published solution. There I saw that I had to square the Möbius transformation to make the image lines perpendicular and finally multiply everything by an appropriate constant to turn the lines to the desired position.

What I don't understand is how you could come up with the idea of squaring the Möbius transformation if you didn't have access to a published solution. If you could somehow realise that the original angle between the image lines is 45 degrees, then it would be natural to use the square function to get a 90 degree angle instead, but it doesn't seem obvious at all that the original angle is precisely 45 degrees and not for instance 43.7 degrees or something else.

RobPratt
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I had it draw your diagram. The angle at which your two circles meet at $i$ or $(0,1)$ is the angle between the tangent lines $y = 1-3x$ and $y = 1 + 2x.$ Or between vectors $(-1,3)$ and $(1,2)$

Meanwhile, the derivative of Moebius transformation $\frac{az+b}{cz+d}$ is simply $\frac{ad-bc}{(cz+d)^2}.$ So, other than the point it sends to infinity, it preserves angles.

enter image description here

Will Jagy
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