Can Mobius Transformations Map Circles Onto Circles?

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Homework Statement



I know that the Mobius transformation:

g(z) = \frac{z-z_1}{z-z_3}\frac{z_2-z_3}{z_2-z_1}

maps a circle (with points z_1, z_2, z_3 somewhere on the circumference) onto a line.

But, i want a general formula for f(z) that maps a circle (z_1,z_2,z_3) ontp a circle (w_1,w_2,w_3)

Homework Equations





The Attempt at a Solution



Does anyone have any ideas?

Thank you in advance
 
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The inverse of g(z) maps a line into a circle; find the general expression for its inverse, select three points in a convinient line and compose the two.
 
I found the inverse. It's z= a function in terms of w and z_1, z_2, z_3.

Why does this help? and what do you mean 'compose the two' ?
 
g(z) takes the points (z1,z2,z3) of the circle to (g(z1),g(z2),g(z3)) in a line. Then choose the inverse f(z) such that it takes the points (g(z1),g(z2),g(z3)) in the line, to (w1,w2,w3) in the circle. The composition f(g(z)) will take (z1,z2,z3) in (w1,w2,w3).
 
the inverse f(z) such that it takes the points (g(z1),g(z2),g(z3)) in the line, to (w1,w2,w3) in the circle is:

f(g(z))=\frac{w-w_1}{w-w_3}\frac{w_2-w_3}{w_2-w_1}

so now i have to combine this with the g(z) is gave in the first post?
Why?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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