Find a Möbius transformation that maps part of a circle to the first quadrant I found this question whilst preparing for an exam:

Let $S = \{ z\in \mathbb{C} | |z-1|<5 , \mathrm{Re}(z)>0 \}$.
Find a Möbius transformation that sends $S$ to the first quadrant $\{ z \in \mathbb{C} | \mathrm{Re}(z) >0 , \mathrm{Im}(z) > 0\}$.

I am puzzled because it seems to me that I have to map the circle and the imaginary axis to the real and imaginary axis, but they make different angles.
 A: With the question as posed, the answer is not possible by Möbius transformations: you are correct. I think that the question should mean "Im>0", rather than "Re>0". The below runs through that calculation.
Consider the map
$$ T(z) = \frac{z+6}{4-z}. $$
Where did I get this from? The two "corners" of the domain are at $z=4$ and $z=-6$. I want to send one to $0$ and one to $\infty$, the "corners" of the region I wish to map to.
Notice that the real axis is a diameter of the circle, therefore they meet at right angles, just as the real and imaginary axes do. Since Möbius transformations are conformal, these angles will be preserved. Further, $T$ clearly maps the real axis (with $\infty$) to itself since the coefficients are all real. Therefore it should map the circle through $4$ and $-6$ at right angles to the real axis to the circle through $0$ and $\infty$ at right angles to the real axis, i.e. the imaginary axis.
Next, I need to check $T$ maps to the right parts of the axes. Clearly for $-6<z<4$ $T(z)>0$, so it maps the diameter to the positive real axis. If $z=-1+5e^{it}$, then
$$ T(z) = \frac{-1+5e^{it}+6}{4+1-e^{it}} = \frac{5(1+e^{it})}{5(1-e^{it})} = \frac{(1+e^{it})(1-e^{-it})}{(1-e^{it})(1-e^{-it})} = \frac{2i\sin{t}}{2(1-\cos{t})}, $$
which is clearly both purely imaginary and with imaginary part larger than zero for $0<t<\pi$.
The last thing to check is that the interior has ended up in the right place. I'm sure you can manage that yourself: just check any one point inside the circle goes to one in the first quadrant. Continuity gives you the rest.
