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.
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.