How to find $T(u) = 0$ given pole of Mobius transformation 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 $T(u) = 0$?

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.
