# How does a conformal mapping preserve angles in hyperbolic geometry?

Suppose I have a sector $D = \{0 < \arg z < \alpha\}$ where $\alpha \leq 2\pi$. If I apply the function $w = \frac{\zeta - i}{\zeta + i}$ from the upper half plane to the unit disc ($\zeta = z^{\frac{\pi}{\alpha}}$), I get that the vertex of the sector goes to -1 and $z = \infty$ goes to 1. I get the unit circle essentially. My question for this example is: How do we know that the angles are preserved?

The answer is that the function sending $\zeta$ to $\frac{\zeta-i}{\zeta+i}$ is holomorphic, and holomorphic functions are those with a complex derivative. If we think of derivatives as being linear maps of best approximation, then that means $f$ is holomorphic if and only if $f(z+w)=f(z)+f^\prime(z)\cdot w+o(w)$. In other words, we are locally just translating and multiplying by $f^\prime (z)$. Translation of course preserves angles, and multiplying by a complex number is the same as a scaling combined with a rotation, which also both preserve angles.
Here's an approach that doesn't require much foundation in complex analysis. The map you have described is among the mappings known as Möbius transformations. These are mappings of the form: $f(z)=(az+b)/(cz+d)$ where $a,b,c,d$ are complex and $ad-bc$ is non-zero.
What you should check is that all Möbius transformations are generated by the following more elementary ones, namely $z\mapsto z+l$, $z\mapsto -\operatorname{conj}(z)$, $z\mapsto kz$ for $k>0$ and finally $z \mapsto 1/\text{conj}(z)$. These you'll notice are all transformations of the upper half plane. In particular, the first two are actually ordinary isometries of the plane. Thusly, they preserve angles. The second to last one only dilates by a constant factor so it, too, preserves angle rather trivially. All that's left to be checked is that the last one preserves angles. You'll notice that this is a composition of $z->-\text{conj}(z)$ and $z->-1/z$. So in fact, we need only check that $z->-1/z$ preserves angles. It is now a not so hard geometric problem to see that this is indeed the case. So in particular, this general argument implies that your specific Möbius transformation preserves angles.