# Show that a holomorphic function $f$ with $f ' \not= 0$ is conformal

Show that a holomorphic function $$f$$ with $$f ' \not= 0$$ is conformal.

I've come across this problem but I couldn't know how to solve it.

I know that a holomorphic function means that it's complex differentiable, meaning that $$\lim_{h\to0} \frac{f(z_0 + h) - f(z_0)}{h}$$ at each $$z_0$$ exists, and is in fact equal to $$f'(z_0)$$. But how would knowing that if $$f'(z_0) \not= 0$$ help me in showing that the function is conformal (i.e. Preserves angles)?

Let $$U$$ be the open set on which $$f$$ is defined. You want to show that for every $$p \in U$$ there exists $$c \in \mathbb R_{>0}$$ such that for all $$v, w \in T_pU \cong \mathbb R^2$$ we have $$\langle (df)_p v, (df)_p w \rangle = c \cdot \langle v, w \rangle$$. That is, that $$(df)_p \in \mathbb C$$, seen as a real $$2 \times 2$$ matrix, lies in the identity component of the general orthogonal group: $$(df)_p \in \operatorname{GO}_2(\mathbb R)^\circ = \left\{ A \in M_2(\mathbb R) : AA^T \in \mathbb R_{>0} \cdot I_2 \right\} \,.$$ But by identifying $$\mathbb C$$ with a subset of $$M_2(\mathbb R)$$, this group is precisely $$\mathbb C^\times$$.
More directly: Let's keep viewing $$T_pU$$ as $$\mathbb C$$ instead of $$\mathbb R^2$$. The inner product on $$T_pU$$ is then given by $$\langle v, w \rangle = \operatorname{Re}(\overline v \cdot w) \,.$$ It is now a triviality that $$f$$ is conformal: \begin{align*} \langle (df)_p v, (df)_p w \rangle &= \langle f'(p) v, f'(p) w \rangle \\ &= \operatorname{Re}(\overline{v f'(p)} \cdot f'(p) w) \\ &= \operatorname{Re}(|f'(p)|^2 \cdot \overline v w) \\ &= |f'(p)|^2 \cdot \langle v, w \rangle \,. \end{align*}