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I have been playing around with this and noticed that it worked for a few functions. Does it work generally? And does that mean that if $f(z)$ is entire or analytic, it's conjugate won't be since $f(\bar{z})$ isn't analytic?

Edi Madi
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    This doesn't hold generally, consider $f(z)=i$ identically. It is true however if $f(x)\in\Bbb R$ for $x\in\Bbb R$. – Wojowu May 02 '16 at 06:45
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    @Wojowu ... and holomorphic (consider $f(z) = \operatorname{Im}(z)$.) – kennytm May 02 '16 at 06:47
  • @kennytm Yes, sure enough. – Wojowu May 02 '16 at 06:50
  • @wojowu, what if I specify that $f(z)$ is not a constant function? – Edi Madi May 02 '16 at 06:52
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    $f(z)=iz{}{}{}$ – Wojowu May 02 '16 at 06:53
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    ...see also http://math.stackexchange.com/questions/102885/how-do-i-rigorously-show-fz-is-analytic-if-and-only-if-overlinef-barz , http://math.stackexchange.com/questions/1644483/analyticity-of-overline-f-bar-z-given-fz-is-analytic?rq=1 , http://math.stackexchange.com/questions/551458/complex-conjugates-of-holomorphic-functions and http://math.stackexchange.com/questions/1617232/to-prove-overline-f-barz-is-analytic?rq=1 – Winther May 02 '16 at 07:04
  • So from what I read in the other threads, if $f(z)$ is analytic, $f(\overline{z})$ and $\overline{f(z)}$ are analytic only when $f(z)$ is constant but $\overline{f(\bar{z})}$ is analytic in general? – Edi Madi May 02 '16 at 07:17

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