- Show that if \(f\) is holomorphic on a connected region \(\Omega\) and \(f'\equiv 0\) on \(\Omega\), then \(f\) is constant on \(\Omega\).
Solution
Goal: Show that a holomorphic function \(f\) on a connected open set \(\Omega\) with \(f' \equiv 0\) is constant.
1. \(\Omega\) is polygonally connected: any two points of \(\Omega\) can be joined by a polygonal path inside \(\Omega\).
\(\Omega\) is an open connected subset of \(\CC\); such sets are path-connected, and the connecting path can be taken polygonal (cover the path by small disks in \(\Omega\) and join centers by segments).
2. If \(f' \equiv 0\) on a segment, then \(f\) is constant along that segment.
Parametrize the segment \(z(t) = z_0 + t(z_1 - z_0)\), \(t \in [0,1]\).
Then \(\ddd{t} f(z(t)) = f'(z(t)) z'(t) = 0 \cdot (z_1 - z_0) = 0\), so \(t \mapsto f(z(t))\) is constant, and \(f(z_1) = f(z_0)\).
3. \(f\) is constant on each convex neighborhood, in particular on each small disk in \(\Omega\).
A disk is convex, so by 2 every point of the disk has the same value as its center (join by the straight segment, which stays in the disk).
4. \(f\) is constant on \(\Omega\).
Fix \(z_0 \in \Omega\).
For any \(z \in \Omega\), join \(z_0\) to \(z\) by a polygonal path (by <1>1); by 2 the value of \(f\) is unchanged along each segment of the path, so \(f(z) = f(z_0)\).
5. Q.E.D.
4 shows \(f\) is constant on the connected region \(\Omega\).