Derivative of Real Valued Complex Function Is Zero
Theorem
Let \(f : U \to \mathbb{C}\) be a differentiable function for \(U \subseteq \mathbb{C}\).
If \(f(z) \in \mathbb{R}\) for all \(z \in U\) then \(f'(z) = 0\) for all \(z \in U\) and hence \(f\) is a constant function.
Proof
Write \(f(x + iy) = u(x, y) + i v(x, y)\) and notice that because \(f(x + iy) \in \mathbb{R}\), we have that \(v(x, y) = 0\) and subsequently
\[ \frac{\partial v}{\partial x} = \frac{\partial v}{\partial y} = 0.\]
By the Cauchy-Riemann equations we also then have that
\[ \frac{\partial u}{\partial x} = \frac{\partial u}{\partial y} = 0\]
and subsequently that \(f' = 0\).
\(f\) must then be the constant function by zero derivative implies constant complex function.