Zero Derivative implies Constant Complex Function

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Theorem

Let $D \subseteq \C$ be a connected domain of $\C$.

Let $f: D \to \C$ be a complex-differentiable function.

For all $z \in D$, let $\map {f'} z = 0$.


Then $f$ is constant on $D$.


Proof

Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ be the two real-valued functions defined in the Cauchy-Riemann Equations:

$\map u {x, y} = \map \Re {\map f z}$
$\map v {x, y} = \map \Im {\map f z}$


By the Cauchy-Riemann Equations:

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

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 $\map u {x + t, y} = \map u {x, y}$ for all $t \in \R$.

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:

$(1): \quad \map \gamma t = z_0 + t r$
$(2): \quad \map \gamma t = z_0 + i t r$

for some $z_0 \in D$ and $r \in \R$ for all $t \in \closedint 0 1$.

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

\(\ds \map f {z_1}\) \(=\) \(\ds \map u {z_0 + t r} + \map v {z_0 + t r}\) for some $t \in \closedint 0 1$
\(\ds \) \(=\) \(\ds \map u {z_0} + \map v {z_0}\) Zero Derivative implies Constant Function
\(\ds \) \(=\) \(\ds \map f {z_0}\)

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

\(\ds \map f {z_1}\) \(=\) \(\ds \map u {z_0 + i t r} + \map v {z_0 + i t r}\) for some $t \in \closedint 0 1$
\(\ds \) \(=\) \(\ds \map f {z_0}\) Zero Derivative implies Constant Function

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 $C$:

$\map f {z_0} = \map f {z_1}$

In particular:

$\map f z = \map f w$

so $f$ is constant on $D$.

$\blacksquare$


Sources

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