# Möbius transformations mapping non-unit circle to non-unit circle

I have a problem in which I need to find a möbius transformation which has as one of the criterion to map the circle $$|z−2+i| = \sqrt5$$ onto the circle $$|w+2| = 2$$, I dont really understand how to extract any information on this about the nature of the transformation.

I do know the other two points, which are $$f(0) = 0$$ and $$f(1-i) = -2$$

If I remember correctly, there is always a mobius transformation that maps any three points to any three points. Moreover, they map circlines to circlines.

So pick three points on the first circle (three points uniquely define a circle by the way), say

$$z_1 = 0 \qquad z_2 = 4 \qquad z_3 = -2i$$

and pick three points on the second circle, say

$$w_1 = 0 \qquad w_2 = -4 \qquad w_3 = -2+2i$$

Suppose we demand that $$f(z_i) = w_i$$ for $$i=1,2,3$$. Since $$f$$ is a mobius transformation, we can let

$$f(z) = \frac{az+b}{cz+d}$$

WLOG, we can set $$a=1$$. Then

\begin{align} f(z_1) = w_1 & \implies \frac bd = 0 \\ f(z_2) = w_2 & \implies \frac{4+b}{4c+d} = -4 \\ f(z_3) = w_3 & \implies \frac{-2i+b}{-2ic+d} = -2+2i \end{align}

Solve for $$b,c,d$$.

• I assume you mean $w_2 = -4$ and $w_3 = -2 - 2i$ or something, but regardless, when I solve that möbius transform, how can I make that circle mapping hold while I also make the third criteria, $f(1-i) = -2$ hold? – acoxy Apr 15 at 16:47
• Why should $f(1-i)=-2$ have to hold? – glowstonetrees Apr 15 at 18:42

Use the fact that the images of those points which are conjugate wrt $$\mathcal C_1$$ will be conjugate wrt $$\mathcal C_2$$. The conjugate of $$1 - i$$ wrt $$\mathcal C_1$$ is $$-3 - i$$, therefore $$f(-3 - i) = \infty$$ and the transform necessarily has the form $$f(z) = a z/(z + 3 + i)$$. Then find $$a$$ from $$f(1 - i) = -2$$.