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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 $f(1-i) = -2$

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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 $b,c,d$.

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  • $\begingroup$ 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? $\endgroup$ – acoxy Apr 15 at 16:47
  • $\begingroup$ Why should $f(1-i)=-2$ have to hold? $\endgroup$ – glowstonetrees Apr 15 at 18:42
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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 $f(1 - i) = -2$.

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