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may i ask you for a little help about the following problem.

Given is the function $f:\mathbb{C}\rightarrow \mathbb{C}, z\rightarrow Re(z)$. The questions are:

1) Does $f$ have an antiderivative on $\mathbb{C}$?

Here i think the answer is no, because the Cauchy-Riemann differential equations are not satisfied, $Re(z)$ is nowhere complex differentiable, thus not holomorphic.

2) Does $f$ have an antiderivative locally on $\mathbb{C}$?

Here i am confused...According to the definition for an antiderivative locally, we need to show that for every $z_{0}\in \mathbb{C}$ there exist an open neighbourhood $V\subset \mathbb{C}$ with $z_{0}\in V$ s.t $f$ on $V$ has an antiderivative. But $f$ is nowhere complex differentiable?

Thank you in advance!

Lord_Farin
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Lullaby
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    I think, you are right. It can't have local antiderivative $F$ on an open $V$, because then $F'=f$ on $V$, so again, also $f$ would be holomorphic on $V$, but it doesn't satisfy C-R in any point. – Berci Apr 04 '13 at 11:38

1 Answers1

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As Berci said, the answer is no to both questions. A function that has an antiderivative in the complex sense, even locally, is necessarily holomorphic -- and $\operatorname{Re}z$ is not holomorphic. ($\LaTeX$ remark: \operatorname{Re}z looks better than Re(z).)

ˈjuː.zɚ79365
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