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Suppose that $T$ is a Mobius transformation that maps the unit circle unto itself. Given that the pole of the map is $2 + i$ (So, $T(2 + i) = ∞$). Find a point $u$ that gets mapped to zero (So, $T(u) = 0$).

I know that Mobius transformation can be defined by three points, which I used as $-1 \, \text{and} \, 1$ on the unit circle, and then use the fact that the denominator will equal 0. So I get something along the lines of

$T(z) = \frac{az + b}{z - (2 + i)}$

Am I using the correct approach? How do I use this to find $T(u) = 0$?

  • There exists a characterization of coefficients $a,b,c,d$ such that the Möbius transformation $z\mapsto \frac{az*b}{cz+d}$ sends the unit circle $|z|=1$ into itself. You are right about the constraint on $a,b,c,d$ given by the pole condition. – Giulio R Oct 01 '22 at 08:17

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A Möbius transformation mapping the unit disc to itself is a Poincaré disc transformation and has the form $$T(z)=e^{i\phi}\frac{z+b}{\bar bz+1}$$ Since here $\bar b(2+i)+1=0$, $b=\overline{-1/(2+i)}=-\frac25-\frac15i$ and the point sent to zero is $-b=\frac25+\frac15i$.

However the transformation here is only required to map the unit circle to itself. Thus the inverse form is also possible: $$T(z)=e^{i\phi}\frac{\bar bz+1}{z+b}$$ But we get the same result for the root of $T$ in this case.

Parcly Taxel
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    Since $T$ is only required to map the unit circle to itself, a second solution is $T(z)=e^{-i\phi}\frac{(2-i)z+1}{z-(2+i)}$, with the same point sent to $0$. – Anne Bauval Oct 01 '22 at 09:03