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Assume that $f$ is analytic in a region and that at every point, either $f\,'= 0$ or $f = 0$. Show that $f$ is constant.

My attempt:

$[f^{2}(z)]\,'=2f(z)f\,'(z)≡0$, so it would only be necessary to clear depending on the condition given

Is my reasoning correct?

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    Note that you still have to show that if $f^2$ is constant, then $f$ is constant... not that hard but it is still a step that needs justification.2012-05-08

1 Answers 1

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You gave one good method. Pedro gives another in the comments: The region is the union of the zero sets of $f$ and $f'$, which implies that at least one of these zero sets has an accumulation point in the region.

Here is another way to apply the identity theorem. Suppose there exists $z_0$ such that $f(z_0)\neq 0$. By continuity, there is an open disk where $f$ is nonvanishing. Therefore $f'$ vanishes on this disk, which implies that $f'\equiv 0$ by the identity theorem for analytic functions.