Given a function $f(z)$ is holomorphic on a disc $D(0,R)$, we want to prove that $g(z)= \overline {f(\bar z)}$ is holomorphic.
Below is my proof.
$\lim_{h\to 0} \frac{\overline {f(\overline{z+h})}-\overline{f(\bar z)}}{h} = \lim_{h\to 0} \frac{\overline {f(\overline{z+h})-\overline{f(\bar z)}}}{h} = \overline{\lim_{h\to 0} \frac {f(\overline{z+h})-\overline{f(\bar z)}}{\bar h}} = \overline{f'(\bar z)}$
Since $ f'(\bar z)$ exists, $g(z)= \overline {f(\bar z)}$ is holomorphic.
This question is actually an old one, but it is not duplicated, because I want to ask how to use Cauchy Riemann Equations to solve it. I mean, if I write $f(z)=u(x,y)+iv(x,y)$ and $g(z)=u(x,-y)-iv(x,-y)$, and try to show $g(z)$ is holomorphic by CREs, what should I do? The relationship between partial derivatives of $u(x,y)$ and $u(x,-y)$ is not just addition inverse. So what should I do next?
Also, I will be very grateful if you can help me to check my proof is correct.
So, actually my question is, how do I write $\frac {\partial} {\partial x } u(x,-y),\frac {\partial} {\partial y } u(x,-y)$ ?
Best Answer
If $f(x,y) = u(x,y)+i v(x,y)$ then $g(x,y) = u(x,-y)-i v(x,-y) = a(x,y)+ib(x,y)$.
Then $ {\partial a(x,y) \over \partial x} = {\partial u(x,-y) \over \partial x}$, $ {\partial a(x,y) \over \partial y} = -{\partial u(x,-y) \over \partial y}$, $ {\partial b(x,y) \over \partial x} = -{\partial v(x,-y) \over \partial x}$ and $ {\partial b(x,y) \over \partial y} = {\partial v(x,-y) \over \partial y}$.
Since ${\partial u(x,-y) \over \partial x} = {\partial v(x,-y) \over \partial y}$ and ${\partial u(x,-y) \over \partial y} = - {\partial v(x,-y) \over \partial x}$ you have the desired result.