Zero Derivative implies Constant Complex Function

Theorem
Let $f: D \to \C$ be a complex-differentiable function, where $D \subseteq \C$ is a connected domain.

Suppose that $f' \left({z}\right) = 0$ for all $z \in D$.

Then $f$ is constant on $D$.

Proof
Let $u, v: \left\{ {\left({x, y}\right) \in \R^2 }\, \middle\vert \, {x+iy = z \in D }\right\} \to \R$ be the two real-valued functions defined in the Cauchy-Riemann Equations:


 * $u \left({x, y}\right) = \operatorname{Re} \left({f \left({z}\right) }\right)$


 * $v \left({x, y}\right) = \operatorname{Im} \left({f \left({z}\right) }\right)$

By the Cauchy-Riemann Equations, it follows that:


 * $f' = \dfrac{\partial f}{\partial x} = \dfrac{\partial u}{\partial x} + i \dfrac{\partial v}{\partial x}$
 * $f' = -i \dfrac{\partial f}{\partial y} = \dfrac{\partial v}{\partial y} - i \dfrac{\partial u}{\partial y}$

As $f' = 0$ by assumption, this implies:


 * $0 = \dfrac{\partial u}{\partial x} = \dfrac{\partial u}{\partial y} = \dfrac{\partial v}{\partial x} = \dfrac{\partial v}{\partial y}$

Then Zero Derivative implies Constant Function shows that $u \left({x + t, y}\right) = u \left({x, y}\right)$ for all $t \in \R$, and similar results apply for the other three partial derivatives.

Let $z, w \in D$.

From Connected Domain is Connected by Staircase Contours, it follows that there exists a staircase contour $C$ in $D$ with endpoints $z$ and $w$.

The contour $C$ is a concatenation of directed smooth curves that can be parameterized as line segments on one of these two forms:


 * $(1): \quad \gamma \left({t}\right) = z_0 + tr$
 * $(2): \quad \gamma \left({t}\right) = z_0 + itr$

for some $z_0 \in D$ and $r \in \R$ for all $t \in \left[{0 \,.\,.\, 1}\right]$.

If $z_1 \in D$ lies on the same line segment as $z_0$, it follows that for paramerizations of type $(1)$:

Similarly, for paramerizations of type $(2)$:

As the image of $C$ is connected by these line segments, it follows that for all $z_0$ and $z_1$ in the image of $C$:
 * $f \left({z_0}\right) = f \left({z_1}\right)$

In particular, $f \left({z}\right) = f \left({w}\right)$, so $f$ is constant on $D$.