1. ## Complex Analysis

Question: Let the function $\displaystyle f$ be holomorphic in the open set $\displaystyle G$. Prove that the function $\displaystyle g(z) = \bar{f\left( \bar{z}\right)}$ is holomorphic in the set $\displaystyle G^\ast =\{\bar{z}: z \in G\}$

Side note: the first bar is over the z and the second bar is over the image of f(z)

My idea for the proof:
I want to show that there are first partial derivatives in G* s.t. the partial derivatives are continuous and satisfy the Cauchy Reimann equations
$\displaystyle du \setminus dx=dv\setminus dy$ $\displaystyle du \setminus dy=-dv \setminus dx$

I differentiate this function on G* w.r.t. to x and y to derive the first partial derivatives.

(how i do this is f(x+h,y)-f(x,y) over h and f(x,y+h) -f(x,y) over ih)

However when I do this I get df/dx = du/dx + -i dv/dx
and df/dy = -dv/dy - i du/dy

which implies the function is constant.

Can someone explain to me what I am doing wrong or not seeing?

2. Originally Posted by arsenicbear
Question: Let the function $\displaystyle f$ be holomorphic in the open set $\displaystyle G$. Prove that the function $\displaystyle g(z) = \overline{f\left( \overline{z}\right)}$ is holomorphic in the set $\displaystyle G^\ast =\{\bar{z}: z \in G\}$.

My idea for the proof:
I want to show that there are first partial derivatives in G* s.t. the partial derivatives are continuous and satisfy the Cauchy Riemann equations
To do this by using the Cauchy–Riemann equations, let $\displaystyle z = x+iy$, $\displaystyle f(z) = u(x,y) + iv(x,y)$ and $\displaystyle g(z) = U(x,y) + iV(x,y)$.

Then $\displaystyle g(z) = \overline{f( \overline{z})} = u(x,-y) - iv(x,-y)$.

Therefore $\displaystyle U(x,y) = u(x,-y)$ and $\displaystyle V(x,y) = -v(x,-y)$. Now you can find the partial derivatives of U and V, and check that they satisfy the C–R equations.

3. I think Opalg's solution is correct but requires an advanced theorem; most early in the study of complex analysis have a weaker theorem to work with (easier to prove): if $\displaystyle f(z)$ satisfies the CRE's at $\displaystyle z_0$ and $\displaystyle f_x, f_y$ (derivatives in the x and y direction) are continuous at $\displaystyle z_0$ then $\displaystyle f$ is differentiable at $\displaystyle z_0$. So you probably have to throw in that $\displaystyle \overline{f_x{(\overline{z_0})}}$ and $\displaystyle \overline{f_y{(\overline{z_0})}}$ are continuous as well.

Alternatively, you can directly apply the definition of the derivative. So for your problem, take any $\displaystyle z_0 \in G^*$. Then $\displaystyle \overline{z_0} \in G \Rightarrow f'(\overline{z_0})$ exists. So, $\displaystyle \overline{f'(\overline{z_0})}$ exists (the last step is simply the conjugate of a complex number). Therefore,
$\displaystyle \lim_{z \rightarrow z_0}{\frac{g(z) - g(z_0)}{z - z_0}} = \lim_{z \rightarrow z_0}{\frac{\overline{f(\overline{z})} - \overline{f(\overline{z_0})}}{z - z_0}}$

$\displaystyle = \lim_{z \rightarrow z_0}{\overline{\left( \frac{f(\overline{z}) - f(\overline{z_0})}{\overline{z} - \overline{z_0}}\right)}} =\overline{\left( \lim_{\overline{z} \rightarrow \overline{z_0}}{\frac{f(\overline{z}) - f(\overline{z_0})}{\overline{z} - \overline{z_0}}}\right)} = \overline{f'(\overline{z_0})}$.