Proposition 2.4.1.
If \(I\) is an interval and \(f: I \to \R\) is a real-valued function with \(f'(x)\) defined and equal to \(0\) for all \(x \in I\text{,}\) then there is a constant \(c \in \R\) such that \(f(x) = c\) for all \(x \in I\text{.}\)
Section 2.4 Constant Functions
As a sample application of the definition of the derivative of a complex function, we consider functions that have a derivative of \(0\text{.}\) In a typical calculus course, one of the first applications of the Mean-Value Theorem for real-valued functions (Theorem A.0.2) is to show that if a function has zero derivative everywhere on an interval then it must be constant.
Proof.
The Mean-Value Theorem A.0.2 says that for any \(x, y \in I\text{,}\)
\begin{equation*}
f(y)-f(x) \ = \ f'\bigl(x + a(y-x) \bigr) (y-x)
\end{equation*}
for some \(0 \lt a \lt 1\text{.}\) Now \(f'(x+a(y-x))=0\text{,}\) so the above equation yields \(f(y)=f(x)\text{.}\) Since this is true for any \(x, y \in I\text{,}\) the function \(f\) must be constant on \(I\text{.}\)
We do not (yet) have a complex version of the Mean-Value Theorem, and so we will use a different argument to prove that a complex function whose derivative is always \(0\) must be constant.
Our proof of Proposition 2.4.1 required two key features of the function \(f\text{,}\) both of which are somewhat obviously necessary. The first is that \(f\) be differentiable everywhere in its domain. In fact, if \(f\) is not differentiable everywhere, we can construct functions that have zero derivative almost everywhere but that have infinitely many values in their image.
The second key feature is that the interval \(I\) is connected. It is certainly important for the domain to be connected in both the real and complex cases. For instance, if we define the function \(f: \{ x+iy \in \C : x \ne 0 \} \to
\C\) through
\begin{equation*}
f(z) := \begin{cases}1\amp \text{ if } \Re z > 0, \\ 2\amp
\text{ if } \Re z \lt 0, \end{cases}
\end{equation*}
then \(f'(z)=0\) for all \(z\) in the domain of \(f\text{,}\) but \(f\) is not constant. This may seem like a silly example, but it illustrates a pitfall to proving a function is constant that we must be careful of. Recall that a region of \(\C\) is an open connected subset.
Theorem 2.4.2.
If \(G\subseteq\C\) is a region and \(f: G \to \C\) is a complex-valued function with \(f'(z)\) defined and equal to \(0\) for all \(z \in G\text{,}\) then \(f\) is constant.
Proof.
We will first show that \(f\) is constant along horizontal segments and along vertical segments in \(G\text{.}\)
Suppose that \(H\) is a horizontal line segment in \(G\text{.}\) Thus there is some number \(y_0 \in \R\) such that the imaginary part of any \(z \in H\) is \(\Im(z) = y_0\text{.}\) Now consider the real part \(u(z)\) of the function \(f(z)\text{,}\) for \(z \in H\text{.}\) Since \(\Im(z) = y_0\) is constant on \(H\text{,}\) we can consider \(u(z) = u(x,y_0)\) to be just a function of \(x\text{,}\) the real part of \(z = x + iy_0\text{.}\) By assumption, \(f'(z) = 0\text{,}\) so for \(z \in H\) we have \(u_x(z) = \Re(f'(z)) = 0\text{.}\) Thus, by Proposition 2.4.1, \(u(z)\) is constant on \(H\text{.}\)
We can argue the same way to see that the imaginary part \(v(z)\) of \(f(z)\) is constant on \(H\text{,}\) since \(v_x(z) = \Im(f'(z))=0\) on \(H\text{.}\) Since both the real and imaginary parts of \(f(z)\) are constant on \(H\text{,}\) the function \(f(z)\) itself is constant on \(H\text{.}\)
This same argument works for vertical segments, interchanging the roles of the real and imaginary parts. We have thus proved that \(f\) is constant along horizontal segments and along vertical segments in \(G\text{.}\) Now if \(x\) and \(y\) are two points in \(G\) that can be connected by a path composed of horizontal and vertical segments, we conclude that \(f(x) = f(y)\text{.}\) But any two points of a region may be connected by finitely many such segments by Theorem 1.4.16, so \(f\) has the same value at any two points of \(G\text{,}\) thus proving the theorem.
There are a number of surprising applications of Theorem 2.4.2; see, e.g., Exercise 2.5.21 and Exercise 2.5.22.
