Complex Analysis - Showing function is holomorphic

Question

I have the following problem:

Suppose $f$ is holomorphic in a region $U^+$. Define $U^{-}: = \{ z: \overline{z} \in U^+\}$. Prove that $g: U^{-} \rightarrow \mathbb{C} $ given by the formula $g(z) = \overline{f(\overline{z})}$ is holomorphic and give $g'$ in terms of $f$.

I don't really know how to start this problem.

I have tried writing out the limit definition for $f$ and then the limit definition for $g$, but getting from $\overline{f(\overline{z})}$ to $f(z)$ doesn't seem easy.

Some help or tips would be appreciated.

Aside: Does anyone know any good sources of problems like this (ideally with solutions)? I have an exam on complex analysis tomorrow so could do with the practice.

Answer 1

There are many ways you can prove this. For example, you can use Cauchy-Riemann equations plus continuity of derivatives, or you can use limit definition, or you can use more advanced tools such as Morera's theorem.

For example, with the limit definition, $$ \frac{g(z)-g(w)}{z-w}=\frac{\overline{f(\bar{z})}-\overline{f(\bar{w})}}{\overline{\bar{z}-\bar{w}}}=\overline{\left(\frac{f(\bar{z})-f(\bar{w})}{\bar{z}-\bar{w}}\right)} $$ which tends to $\overline{f'(\bar{z})}$ as $w\to z$ in $U^-$ (equivalently $\bar{w}\to\bar{z}$ in $U^+$).

Answer 2

Hint: If $f(x+iy) = u_1(x,y)+i v_1(x,y)$ and $g(x+iy) = u_2(x,y) + i v_2(x,y)$, using $g(z) = \overline{f(\overline z)}$ express $u_2$ and $v_2$ in terms of $u_1$ and $v_1$. Now check the Cauchy-Riemann equations.