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A holomorphic function with zero derivative on a domain is constant

Statement

Let $U \subseteq C$ be a domain. If $f : U \to C$ is holomorphic and $f^{'} (z) = 0$ for every $z \in U$ , then $f$ is constant on $U$ .

Facts & Assumptions

Given: A complex domain $U$ and a holomorphic function $f : U \to C$ with $f^{'} = 0$ throughout $U$ .

[F1]

A complex domain is a nonempty connected open subset of $C$ (A complex domain is a nonempty connected open subset of $C$ ).

[L1]

For an open subset of $R^{n}$ , connectedness, path-connectedness, and polygonal connectedness are equivalent (For an open subset of $R^{n}$ , connectedness, path-connectedness and polygonal connectedness are equivalent).

[F2]

A polygonal path is a finite concatenation of affine line segments with consecutive vertices (Polygonal paths and polygonally connected subsets of $R^{n}$ ).

[L2]

If $f$ is complex differentiable at a point, then its real total derivative is multiplication by $f^{'}$ (Complex differentiability is equivalent to real total differentiability together with a complex-linear derivative, with $\partial_{\overset{ˉ}{z}} f = 0$ , or with the Cauchy–Riemann equations).

[L3]

The total derivative of a composite is the composite of the total derivatives (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

[L4]

A complex-differentiable function is continuous at the point of differentiability (Complex differentiability at a point implies continuity there).

[L5]

A real function continuous on an order-convex interval and differentiable at every interior point, with derivative zero there, is constant on that interval (A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant).

Proof

technique · direct

1.1

Let $p, q \in U$ . By [F1] and [L1], choose a polygonal path in $U$ from $p$ to $q$ , with vertices $v_{0} = p, v_{1}, \dots, v_{m} = q$ as in [F2].

givenF1L1F2choose

2.1

For each segment put $γ_{j} (t) = (1 - t) v_{j - 1} + t v_{j}$ for $0 \leq t \leq 1$ . The composite $f \circ γ_{j}$ is continuous on $[0, 1]$ by [L4], including at the endpoints.

step 1.1L4

2.2

At every $t \in (0, 1)$ , [L2] makes $D f (γ_{j} (t))$ multiplication by $0$ , and [L3] therefore gives $D (f \circ γ_{j}) (t) = 0$ . Hence the real and imaginary components of $f \circ γ_{j}$ have derivative zero on $(0, 1)$ .

step 1.1givenL2L3algebra

3.1

By [L5], both components are constant on $[0, 1]$ , so $f (v_{j - 1}) = f (v_{j})$ for each segment, including a zero-length segment if one occurs.

step 2.1step 2.2L5

4.1

Chaining the finitely many equalities from step 3.1 gives $f (p) = f (q)$ . Since $p, q$ were arbitrary, $f$ is constant on the nonempty domain $U$ .

step 1.1step 3.1F1∎

Depends on

Used by

Dependency tree · two levels

35 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Guide to Cultivating Complex Analysis, Proposition 2.2.1 (standard reference, not scraped)
R. Howell and J. Mathews, Complex Analysis, Theorem 3.2.13 (standard reference, not scraped)