Let $f = u + iv$ be holomorphic and $g= w + il$ be antiholomorphic (that is $w -il$ is holomorphic) in $\Omega \subset \mathbb{C}$. What can we say about their product $f\cdot g$?
Product of holomorphic and antiholomorphic functions
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calculus
complex-analysis
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1Do you have any particular properties in mind...? – 2017-01-06
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1@AndrewD.Hwang Holomorphic, antiholomorphic, analytic, harmonic, a polynomial, ... – 2017-01-06
1 Answers
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By considering $z \mapsto z \overline z$ you can see that you have no chances to have holomorphic, anti-holomorphic, harmonic or polynomial functions. You only have $C^{\infty}$ function since both holomorphic and anti-holomorphic functions are $C^{\infty}$.
Remark : If you consider $f(z) = \sum a_n z^n$ and $g(\overline z) = \sum b_n \overline z^n$ you can notice that the product $f(z)g(\overline z)$ is real analytic considered as a function of $(x,y)$. This is also $C^{\infty}$ as a function of $(z, \overline z)$ and you can't say anything more interesting in general.
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0What do you mean when you say "By considering $z \mapsto z \overline z$ you can see ..."? – 2017-01-06
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1$z \to z$ is holomorphic and $z \to \overline z$ is anti-holomorphic. And $z \overline z = |z|^2$ is not holomorphic, etc ... You can consider it as a "counter-example" to what you did expected in comments. – 2017-01-06
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0@Axel : I did put a little remark. I don't think you can say anything more in general. – 2017-01-06