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    Zeros of $f'$ are isolated, and $f'=g'$ implies $f-g$ is constant

    PR-NZZ2C

    A nonconstant holomorphic function has isolated critical points, and equal derivatives differ only by a constant.Suggest a slogan

    Let \(\Omega\subseteq\CC\) be a connected open set.

    1. If \(f\) is analytic and nonconstant on \(\Omega\), then \(f'\) is analytic on \(\Omega\) and the zeros of \(f'\) are isolated.

    2. If \(f,g\) are analytic on \(\Omega\) with \(f'=g'\), then \(f-g\) is constant on \(\Omega\).

    1. A power series can be differentiated term by term inside its disc of convergence, so \(f'\) is analytic on \(\Omega\). If \(f'\equiv0\) on \(\Omega\), then \(f\) is locally constant, hence constant because \(\Omega\) is connected. So \(f'\) is an analytic function on a connected open set that is not identically zero, and its zeros are isolated.

    2. Put \(h\coloneqq f-g\). Then \(h'=0\) on \(\Omega\), so \(h\) is locally constant: on a disc \(D\subseteq\Omega\), \(h(z)-h(z_0)=\int_{[z_0,z]}h'(w)\,dw=0\). The set \(\ts{z\in\Omega\st h(z)=h(z_0)}\) is therefore open, and it is closed by continuity, so it is all of \(\Omega\).

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