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I took a final examination recently in a complex analysis course, and one of the questions still alludes me. I wrote it down after the exam to work on it later, but I'm not seeing how to do this. It reads as follows:

Consider a Mobius transformation $T$ with real coefficients such that $T(0)=0$ and $T(2)=\infty$. What set is mapped to the imaginary axis under $T$? That is, what is $T^{-1}(\{iy : y \in \mathbb{R}\})$?

Now, I know that under these linear fractional transformations, circles and lines are mapped to circles and lines, and a given mapping can be determined by its action on three points, but I'm not sure how I can apply this here. Obviously, it would seem that the set we're looking for will be either a circle or a line, but I also know that a Mobius transformation with real coefficients (assuming $ad-bc \neq 0$) maps the extended real line to the extended real line. I imagine this will be helpful, but I'm still not catching the trick yet, so I figured I'd extend the problem out for a hint or nudge in the right direction.

My only idea would be perhaps mapping something like a circle with radius 1 centered at $(1, 0)$, but this doesn't quite give us the imaginary axis we're looking for.

Best,
JR

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  • $\begingroup$ Why is your conjecture not right? $\endgroup$ May 13, 2023 at 3:28

2 Answers 2

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$T(z)=\frac{az+b}{cz+d}$ with $a,b,c,d\in\Bbb R$ such that ($ad-bc\ne0$ and) $0=T(0)=\frac bd$ i.e. $b=0,$ and $\infty=T(2)=\frac{2a}{2c+d}$ i.e. $d=-2c,$ so $$T(z)=k\frac z{z-2},\quad k\in\Bbb R^*.$$ For every complex number $z\ne2,$ $$\begin{align}z\in T^{-1}(i\Bbb R)&\iff \overline{T(z)}=-T(z)\\&\iff \frac{\bar z}{\bar z-2}=-\frac z{z-2}\\&\iff|z|^2-z-\bar z=0\\&\iff x^2+y^2-2x=0\\&\iff(x-1)^2+y^2=1, \end{align}$$ so your conjecture was right.

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Let $C$ be the preimage of the extended imaginary axis under $T$. We know that

  • $C$ is a circle or an extended line.
  • $C$ contains the points $T^{-1}(0) = 0$ and $T^{-1}(\infty) = 2$,
  • $C$ is symmetric with respect to the real axis (because $T$ either preserves or exchanges the upper and lower halfplane),
  • $C$ cuts the real axis at $0$ and $2$ in a right angle (because $T$ preserves angles).

The last property excludes that $C$ is a line, so it is a circle through the points $0$ and $2$ which is symmetric with respect to the real line.

It follows that $C$ is the circle with radius $1$, centered at $1$, and $T^{-1}(\{iy : y \in \Bbb R\})$ is that circle without $T^{-1}(\infty) = 2$.

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