Given a Mobius transformation with real coefficients such that $T(0)=0$ and $T(2)=\infty$, what is $T^{-1}(\{iy : y \in \mathbb{R}\})$?

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

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

$$T(z)=\frac{az+b}{cz+d}$$ 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.
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 $$T^{-1}(\infty) = 2$$.