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Please check whether this makes sense

If [itex]U\subset\mathbb{C}[/itex] open, path-connected and [itex]f:U\longrightarrow\mathbb{C}[/itex] differentiable with [itex]f'(z)=0[/itex] for all [itex]z\in{U}[/itex], then f is constant.

Hypotheses:

H1: U is pathconnected

H2: [itex]f:U\longrightarrow\mathbb{C}, f'(z)=0, z\in{U} [/itex]

By H1 for every [itex]p,q\in{U},p\neq{q}[/itex] there exists [itex]\gamma:[a,b]\longrightarrow{U}[/itex] such that [itex]\gamma(a)=p,\gamma(b)=q[/itex].

So let [itex]\gamma[/itex] be an arbitrary path in U and let [itex]z\in\gamma([a,b])\subset{U}[/itex], that is [itex]z:=\gamma(t)[/itex] for [itex]t\in[a,b][/itex].

We shall show that f is constant on any path in U between arbitrary but fixed points p and q. Since [itex]\gamma[/itex] is arbitrary, it will then follow that f is constant on U.

Since f is complex-differentiable and gamma is continuous, and hence also real-differentiable, we have [itex](f\circ{\gamma})'(t)={\gamma}'(t)f'(\gamma(t))[/itex]. But [itex]f'(\gamma(t))=f'(z)=0[/itex] by H2, hence [itex](f\circ{\gamma})'(t)=0[/itex], which implies that [itex](f\circ{\gamma})(t)=const[/itex]. In particular, [itex](f\circ{\gamma})(a)=(f\circ{\gamma})(b)[/itex], that is [itex]f({\gamma}(a))=f({\gamma}(b))[/itex], or [itex]f(p)=f(q)[/itex] for all [itex]p\neq{q}, p,q\in U[/itex]. But this is exactly the property of a constant function. It follows that f is locally, on a subset of U, constant and hence constant.

## Homework Statement

If [itex]U\subset\mathbb{C}[/itex] open, path-connected and [itex]f:U\longrightarrow\mathbb{C}[/itex] differentiable with [itex]f'(z)=0[/itex] for all [itex]z\in{U}[/itex], then f is constant.

Hypotheses:

H1: U is pathconnected

H2: [itex]f:U\longrightarrow\mathbb{C}, f'(z)=0, z\in{U} [/itex]

**2. The attempt at a solution**By H1 for every [itex]p,q\in{U},p\neq{q}[/itex] there exists [itex]\gamma:[a,b]\longrightarrow{U}[/itex] such that [itex]\gamma(a)=p,\gamma(b)=q[/itex].

So let [itex]\gamma[/itex] be an arbitrary path in U and let [itex]z\in\gamma([a,b])\subset{U}[/itex], that is [itex]z:=\gamma(t)[/itex] for [itex]t\in[a,b][/itex].

We shall show that f is constant on any path in U between arbitrary but fixed points p and q. Since [itex]\gamma[/itex] is arbitrary, it will then follow that f is constant on U.

Since f is complex-differentiable and gamma is continuous, and hence also real-differentiable, we have [itex](f\circ{\gamma})'(t)={\gamma}'(t)f'(\gamma(t))[/itex]. But [itex]f'(\gamma(t))=f'(z)=0[/itex] by H2, hence [itex](f\circ{\gamma})'(t)=0[/itex], which implies that [itex](f\circ{\gamma})(t)=const[/itex]. In particular, [itex](f\circ{\gamma})(a)=(f\circ{\gamma})(b)[/itex], that is [itex]f({\gamma}(a))=f({\gamma}(b))[/itex], or [itex]f(p)=f(q)[/itex] for all [itex]p\neq{q}, p,q\in U[/itex]. But this is exactly the property of a constant function. It follows that f is locally, on a subset of U, constant and hence constant.

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