# Prove a function is holomorphic

Let

• $A \subset \mathbb{C}$ be an open set
• $f : A \rightarrow \mathbb{C}$ be a holomorphic function
• $B = \{z \in \mathbb{C} | \bar{z} \in A\}$
• $g(z) : B \rightarrow \mathbb{C} = \overline{f(\bar{z})}$

Show by using the Cauchy-Riemann equations that $g$ is holomorphic for $B$.

I do have problems to understand the Cauchy Riemann equations. As far as I understand I need to show that g is complex differentiable $\forall z \in B$, and therefore the first derivate at $g(z)$ exists, right? The Cauchy-Riemann equations allow to swap some arguments of partial derivatives, but how can this be used here?

• Unless $f$ is constant, $g$ is not holomorphic, it should be $g(z) = \overline{f(\overline{z})}$. – Daniel Fischer Apr 26 '14 at 15:28
• Also, once you change the problem, as Daniel mentioned, you should be able to use the chain rule to show that the Cauchy Riemann equations for $\overline{f(\overline{z})}$ and $f(z)$ are equivalent. – Siddharth Venkatesh Apr 26 '14 at 15:40
• @DanielFischer thank you, you were right! I fixed it in the question – muffel Apr 26 '14 at 15:41

Write $$f(x,y)=u(x,y)+i v(x,y),\quad g(\xi,\eta)=a(\xi,\eta)+ib(\xi,\eta)$$ where $u$, $v$, $a$, $b$ are realvalued functions defined in $A$, resp. $B$. Then by definition of $g$ one has $$a(\xi,\eta)+ib(\xi,\eta)=g(\xi+i\eta)=\overline{f(\xi-i\eta)}=u(\xi,-\eta)-iv(\xi,-\eta)$$ and therefore $$a(\xi,\eta)=u(\xi,-\eta),\quad b(\xi,\eta)=-v(\xi,-\eta)\qquad\bigl((\xi,\eta)\in B\bigr)\ .$$ It follows that, e.g. $$b_\eta(\xi,\eta)=-v_y(\xi,-\eta)\cdot(-1)=v_y(\xi,-\eta)\ .$$ Since $u$ and $v$ satisfy the CR-equations in the variables $x$ and $y$ we conclude that $$a_\xi(\xi,\eta)=u_x(\xi,-\eta)=v_y(\xi,-\eta)=b_\eta(\xi,\eta)\ ,$$ and similarly $$a_\eta(\xi,\eta)=-u_y(\xi,-\eta)=v_x(\xi,-\eta)=-b_\xi(\xi,\eta)\ .$$ This shows that $g$ fulfills the CR-equations in the variables $\xi$ and $\eta$.

But there is also a direct approach, which in my view is simpler and more in tune with a complex world description.

As $f$ is holomorphic in $A$, for each point $z_0\in A$ (held fixed in the following) there is a complex number $C$ such that $$f(z)-f(z_0)=C(z-z_0)+o(|z-z_0|)\qquad (z\in A, \ z\to z_0)\ .$$ Let a point $w_0\in B$ be given, and put $z_0:=\bar w_0$. Then by definition of $g$ one has $$g(w)-g(w_0)=\overline{f(\bar w)}-\overline{f(\bar w_0)}=\overline{f(\bar w)-f( z_0)}=\overline{C(\bar w -z_0)+o(|\bar w-z_0|)}\qquad(w\in B)\ .$$ As $|\bar w -z_0|=|w-w_0|$ it follows that $$g(w)-g(w_0)=\bar C(w-w_0)+o(|w-w_0|)\qquad(w\in B, \ w\to w_0)\ .$$ It follows that $g'(w_0)=\bar C$, and as $w_0\in B$ was arbitrary, we conclude that $g$ is holomorphic in $B$.

• thank you for your answer. Can you please explain to me what exactly the difference between $a, a_x$ and $a_y$ is? – muffel Apr 28 '14 at 11:17
• @muffel: See my edit. With $u_x$, $a_\xi$, etc. one denotes the partial derivative of the functions $u$, $a$, etc. with respect to the indicated variable. – Christian Blatter Apr 28 '14 at 11:43
• Arf, I didn't see the second part of this answer before posting mine :/. Let's mention that to apply Cauchy-Riemann, we need to check that $g$ is differentiable. – Taladris Apr 28 '14 at 12:58
• @ChristianBlatter great, thank you very much! – muffel Apr 28 '14 at 14:31

You can also prove it using the definition. Everything works nicely since the conjugation $z\mapsto\overline{z}$ is continuous.

For any $z_0$ in $B$, we have $\frac{g(z)-g(z_0)}{z-z_0}=\frac{\overline{f(\overline{z}})-\overline{f(\overline{z_0})}}{z-z_0}=\overline{\ \frac{f(\overline{z})-f(\overline{z_0})}{\overline{z}-\overline{z_0}}\ }$.

But, by continuity of the conjugation, $z\to z_0$ if and only if $\overline{z}\to\overline{z_0}$, so $\displaystyle{ \lim_{z\to z_0} \frac{g(z)-g(z_0)}{z-z_0} = \overline{ \lim_{\overline{z}\to\overline{z_0}} \frac{ f(\overline{z})-f(\overline{z_0}) }{ \overline{z}-\overline{z_0}} } =\overline{f'(\overline{z_0})} }$

(the last equality justified by the fact that $\overline{z_0}$ belongs to $A$).