Let $f: D \rightarrow \mathbb{C}$ be defined on open connected $D \subset \Bbb{C}$ and be differentiable with $f'(z)=0$ for all $z \in D$. Show $f$ is constant.
Proof.
Fix $z_0 \in D$ and let $f(z_0)=w_0$ and set
$$A=\{z \in D : f(z)=w_0\}$$
We show since $D$ is connected, if $A \subset D$ is both open and closed then it must be all of $D$ forcing $f$ to be constant on all of $D$. To show $A \subset D$ is closed (which is the easier direction), let $z \in D$ and $\{z_n\}\subset A$ with
$$\lim_{n \rightarrow \infty} z_n=z$$
Then as $z_n \in A$ for all $n \geq 1$, it follows that $f(z_n)=w_0$ and by continuity of $f$ we have that $f(z)=w_0$ that is, $z \in A$ thus $A$ contains its own limit points and is thus closed in $D$. For open, fix some $a \in A$ and since $D$ is open we can find an $\epsilon >0$ such that
$$B_\epsilon(a) \subset D.$$
Then for $z \in B_\epsilon(a)$ define
$$g(t):=f(tz+(1-t)a)$$
for $t \in [0,1]$, this is made possible by path-connectedness of $B_\epsilon(a)$. (right?)
Then we have that, by the chain rule
$$\lim_{t \rightarrow s} \frac{g(t)-g(s)}{t-s}=f'(sz+(1-s)a)(z-a)$$
But $sz+(1-s)a \in D$ thus $f'$ here is zero forcing
$$g'(s)=0 \space \text{for all $s \in [0,1]$}$$
and so $g$ is constant. Thus $f(z)=g(1)=g(0)=f(a)=w_0$ hence $B_\epsilon(a) \subset A$ since $f(z),f(a)=w_0$ and $A$ is open as needed. $\blacksquare$