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)? 
 A: 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:
$$\begin{align*}
\langle (df)_p v, (df)_p w \rangle
&= \langle f'(p) v, f'(p) w \rangle \\
&= \operatorname{Re}(\overline{v f'(p)} \cdot f'(p) w) \\
&= \operatorname{Re}(|f'(p)|^2 \cdot \overline v w) \\
&= |f'(p)|^2 \cdot \langle v, w \rangle \,.
\end{align*}$$
