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If $f$ is an entire function then is $f(\bar{z})$ is also entire?

I'm thinking about it a lot and I think changing the variable from $z$ to $\bar{z}$ doesn't change anything because it's already analytic on every point of $\mathbb{C}$.

AmI correct?

Itachi
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    is $\bar{z}$ entire? – Sean Nemetz Jul 10 '22 at 19:18
  • @SeanNemetz sorry? – Itachi Jul 10 '22 at 19:19
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    Just consider $f$ the identity. – Randall Jul 10 '22 at 19:19
  • is the function $f(\bar{z})=\bar{z}$ analytic everywhere on the complex plane? $z$ is definitely entire, however... – Sean Nemetz Jul 10 '22 at 19:20
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    @SeanNemetz can you please elaborate that why $\bar{z}$ is not analytic everywhere? – Itachi Jul 10 '22 at 19:26
  • Use Cauchy-Riemann and the chain rule for Wirtinger derivatives: $$\frac{\partial}{\partial \bar z}f(\bar z)=\frac{\partial f}{\partial z}(\bar z)\frac{\partial \bar z}{\partial \bar z}+\frac{\partial f}{\partial \bar z}(\bar z)\frac{\partial z}{\partial \bar z}=f'(\bar z)$$ Therefore given $f$ holomorphic on some open set, $f(\bar z)$ is holomorphic if and only if $f$ is locally constant on its domain. – Sassatelli Giulio Jul 10 '22 at 19:34
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    Check this: https://math.stackexchange.com/q/1138819/42969 – Martin R Jul 10 '22 at 19:36
  • @SassatelliGiulio Thank you but I'm afraid I don't understand this because I don't know "Wirtinger derivative" yet. – Itachi Jul 10 '22 at 19:42
  • @Itachi Ok, I'll keep in mind not to delete the comment, then. – Sassatelli Giulio Jul 10 '22 at 19:46
  • @Itachi You are probably missing a bar.$\overline{f(\overline{z})} $ – Sourav Ghosh Jul 10 '22 at 19:49
  • Differentiation free approach: assume WLOG $f(0)=0$. If $f(\bar z)$ is analytic then $f(z)=f(\bar z)$ (why?). Consider closed disc $\overline B(0,r)$ with $r$ small enough so $f$ is non-zero on the punctured disc $\overline B^*(0,r)$. Let closed curve $\gamma:[0,1]\longrightarrow \mathbb C$ be given by $\gamma(t)=r\cdot \exp(2 \pi i \cdot t)$. Then for $t\in [\frac{1}{2},1]$ we have $f\circ \gamma(t) = f\circ \gamma(1-t) $ i.e. it is the reversal of $f\circ \gamma(t)$ for $t\in [0,\frac{1}{2}]$. This contradicts Argument Principle for a disc. (This can be made rigorous with winding numbers.) – user8675309 Jul 11 '22 at 05:01

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