Is $g(z)$ holomorphic if $f$ is holomorphic in $G$?

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  • Thread starter Euge
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In summary, $g(z)$ is not always holomorphic if $f(z)$ is holomorphic in $G$. If $g(z)$ is holomorphic in $G$, it can be expressed as $g(z) = f(z) + C$, where $C$ is a constant. It cannot have poles or essential singularities if $f(z)$ is holomorphic in $G$. However, the converse is not always true, as $g(z)$ could be a composition of different functions. Additionally, $g(z)$ can be a non-analytic function even if $f(z)$ is holomorphic in $G$.
  • #1
Euge
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Here is this week's POTW, suggested by Chris L T521:

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$\newcommand{\CC}{\mathbb{C}}$ Let the function $f$ be holomorphic in the open set $G\subset\CC$. Prove that the function $g(z)=\overline{f(\overline{z})}$ is holomorphic in the set $G^{\ast}=\{\overline{z}:z\in G\}$.
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  • #2
This week's problem was answered correctly by Cbarker1. You can read his solution below.
Let $g: G^{\ast} \to \Bbb{C}$ is given by $g(z)=\overline{f(\overline{z})}$. By definition of holomorphic, let $z_0 \in G^{\ast}$ be arbitrary, we have
\begin{align*}
&g'(z_0)=\lim_{{z}\to{z_0}}\frac{g(z)-g(z_0)}{z-z_0}\\
& =\lim_{{z}\to{z_0}}\frac{\overline{f(\overline{z})}-\overline{f(\overline{z_0})}}{z-z_0} & (\text{by the definition of g})\\
& =\lim_{{z}\to{z_0}}\frac{\overline{f(\overline{z})-f(\overline{z_0})}}{z-z_0} \\
&=\overline{f'(\overline{z_0})} &(\text{f is holomorphic}).
\end{align*}Thus, $g$ is holomorphic in the set $G^{\ast}$.
 

FAQ: Is $g(z)$ holomorphic if $f$ is holomorphic in $G$?

Is it possible for $g(z)$ to be holomorphic if $f$ is not holomorphic in $G$?

No, if $f$ is not holomorphic in $G$, then it is not possible for $g(z)$ to be holomorphic in $G$. In order for $g(z)$ to be holomorphic, it must satisfy the Cauchy-Riemann equations, which require $f$ to be holomorphic in $G$.

Can $g(z)$ be holomorphic in $G$ if $f$ is only holomorphic on a subset of $G$?

Yes, it is possible for $g(z)$ to be holomorphic in $G$ even if $f$ is only holomorphic on a subset of $G$. As long as $g(z)$ satisfies the Cauchy-Riemann equations on that subset, it can be extended to be holomorphic in $G$.

What is the relationship between the holomorphicity of $f$ and $g(z)$?

If $f$ is holomorphic in $G$, then $g(z)$ is also holomorphic in $G$. This is because $g(z)$ is defined as the derivative of $f$ with respect to $z$, and the derivative of a holomorphic function is also holomorphic.

Can $g(z)$ be non-holomorphic in $G$ if $f$ is non-holomorphic in $G$?

Yes, it is possible for $g(z)$ to be non-holomorphic in $G$ even if $f$ is non-holomorphic in $G$. This can happen if $g(z)$ does not satisfy the Cauchy-Riemann equations, even if $f$ does.

Are there any other conditions that need to be met for $g(z)$ to be holomorphic in $G$?

Yes, in addition to $f$ being holomorphic in $G$, $g(z)$ must also be continuous in $G$. This is because holomorphic functions are also continuous. Additionally, $g(z)$ must have a continuous derivative in $G$ in order for it to be holomorphic.

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