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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?

  • 2
    I see no problems.2012-05-08
  • 0
    What if the region is not connected?2012-05-08
  • 1
    @nullUser, [region](http://en.wikipedia.org/wiki/Region_(mathematical_analysis) usually means connected and open.2012-05-08
  • 2
    Another option (more intuitive in my opinion) is to use the fact that if the zeroes of an analytic function accumulate, it's identically zero. By the pidgeonhole principle you will have the zeroes of either $f$ or $f'$ accumulating.2012-05-08
  • 0
    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.