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Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be analytic and suppose that for all $z \in \mathbb{C}$, at least one of $f(z)$ and $f'(z)$ is equal to 0. Proof that $f$ is constant.

Any ideas? Thanks.

Davide Giraudo
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MrReese
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    In your title you mention $f''$, but in your body $f'$. Which do you mean? – Chris Eagle Mar 10 '13 at 20:42
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    $f'$ thanks, fixed it. – MrReese Mar 10 '13 at 20:44
  • We have that $f(z) f'(z) = 0$ So if $f(z)$ is not $0$ then $f'(z) = 0$ However this implies the value does not change in the neighbourhood of $f(z)$. If almost every point is an extremum then $f$ cannot be analytic and unbounded at the same time. Hence $f$ is constant. That is not formal , its just a comment. Notice an entire function only has at most a countable amount of zero's or it is identical to $0$. – mick Mar 10 '13 at 20:48

3 Answers3

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Hint: $$ (?)'(z)=f(z)f'(z)=0\qquad\forall z. $$

Julien
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  • Hehe...+1 for that teasing yet nice (?) there. – DonAntonio Mar 10 '13 at 20:45
  • @julien I may be missing something, but don't you mean $((?)(z))'$ and of course $(?)'(z)$ also works for an appropiate $?$, but if I got it right it $((?)(z))'$ should be clearer. – Git Gud Mar 10 '13 at 20:48
  • @GitGud does not matter since $dz = 1$. – mick Mar 10 '13 at 20:51
  • In case the OP is confused , Integrate ... – mick Mar 10 '13 at 20:52
  • @GitGud I was thinking $(?)=g$ with $g=f^2/2$. I added the parenthesis because I wanted the $'$ to be more apparent than in $?'$. – Julien Mar 10 '13 at 20:57
  • @julien Ok. Nice answer as always anyway. – Git Gud Mar 10 '13 at 20:58
  • @GitGud Thanks a lot. – Julien Mar 10 '13 at 20:59
  • @GitGud: There aren't many contexts where $(g(z))'$ is meaningful. One would be for instance where $g:\mathbf C\to\mathcal C^1(\mathbf R)$, a map whose values are differentiable functions (and the value $(g(z))'$ would be the unevaluated derivative function of the image $g(z)$ of $z$). It is not likely that is useful here. – Marc van Leeuwen Mar 11 '13 at 17:26
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Alternative ... since $\{z:f(z)=0\}$ and $\{z:f'(z)=0\}$ are both closed sets, if their union is the whole plane then at least one of them has nonempty interior.

addition
The same method works for a countable union by the Baire category theorem. So:

Let $f$ be an entire function, and suppose for each $z \in \mathbb C$ at least one of $f(z), f'(z), f''(z), f'''(z),\cdots$ is zero. Then $f$ is a polynomial.

GEdgar
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  • Despite my answer, I like it when one uses a more involved theorem like Baire to solve such a question. Thanks for sharing, +1. And that's a very nice generalization. – Julien Mar 11 '13 at 18:17
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Expanding on the comments:

Since one of $f$ and $f'$ is zero everywhere, for all $z$, $0 = f(z)f'(z) = (1/2)(f(z)^2)'$, so $f(z)^2$ is constant, so $f(z)$ is constant.

marty cohen
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