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What does it mean that $$w=\frac{az+b}{cz+d}$$ can map a circle to a line, a line to a line, a circle to a circle or a line to a circle?

Thanks for your explanation. I noticed many problems regarding Conformal Mapping are asking to form a map in order to get a desired result, for instance, fix some points or map a area to another one. But I don't know how can I find the desired mapping. Is there a certain rules I should follow? The following is an example. Let U be the upper half plane from which the points of the

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is $a,b\in\mathbb{R}$ or $a,b\in\mathbb{C}$ ? –  Abdelmajid Khadari Apr 25 '12 at 1:08
    
Here is a link to a very enlightening video. It may help build your intuition about what is behind these mappings. youtube.com/watch?v=JX3VmDgiFnY –  John Adamski Apr 25 '12 at 3:39
    
@ John It truly is very enlightening an beautiful. Thank you. –  Megan Apr 25 '12 at 18:49

1 Answer 1

It does not really matter about the coefficients. Any (invertible) transformation has $ad-bc \neq 0.$ Such a mapping, written as a function, is the composition of a finite string of just three types:

First $f(z) = A z$ for some nonzero (real or complex) $A$ takes circles with center at $0$ to others, lines through $0$ to others.

Second $g(z) = z + B$ takes lines to lines and circles to circles, it is just a translation.

Third, $h(z) = \frac{-1}{z}$ does a little of both. Since $0$ is sent out to $\infty$ and $\infty$ is brought back to $0,$ the following bunch of things happen:

A) a line through $0$ is sent to a line through $0.$

B) a line not through $0$ is sent to a circle that passes through $0,$ with center elsewhere.

C) A circle that passes through $0$ is sent to a line that does not pass through $0.$

D) A circle that does not pass through $0$ is sent to another circle that does not pass through $0.$

You really ought to draw some things yourself for $h(z) = -1/z.$ There is a good reason for the minus sign, but if you prefer draw $H(z) = 1/z.$

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Thanks for your explanation. –  Megan Apr 25 '12 at 1:48

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