# 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?

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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.