$E$ be a domain, define $E^=\{z \in C: \overline{z}\in E\}$ $f: E \to C$ is analytic, then $f^(z)=\overline{f(\overline{z})}$ analytic on $E*$.

Question

We know that since $f$ is analytic on $E$ we have

$$u_x=v_y \quad u_y=-v_x$$

We have $f^*(z)=u(x,-y)-iv(x,-y)$

Essentially we are going from

$$E^* \stackrel{\overline{z}}{\rightarrow} E \stackrel{f}{\rightarrow} \mathbb{C} \stackrel{\overline{z}}{\rightarrow} \mathbb{C}$$

and $f$ is analytic on $E$. While $z \mapsto \overline{z}$ is not analytic we are doing that mapping twice so it may not disturb analyticity?

I know I need to consider the Cauchy-Riemann equations on $f^*$ but I'm not sure how to get there. Thanks for the help!

It is a composition of an antiholomorphic map, with a holomorphic map, with an antiholomorphic map again. — Future, Mar 05 '16 at 23:14
Well, it shows that the function is holomorphic. What else are you asking? — Future, Mar 06 '16 at 03:52
The composition of two antilinear maps will be linear. Apply it to $\mathbf{C}$-linear maps. — Future, Mar 06 '16 at 03:56

score 2 · Accepted Answer · answered Mar 06 '16 at 05:17

Let $U(x,y)=\operatorname{Re}\, f^*(z), V(x,y)=\operatorname{Im}\, f^*(z).$ Then \begin{align} &U(x,y)=u(x,-y),\quad V(x,y)=-v(x,-y),\\ &f^*(z)=U(x,y)+iV(x,y). \end{align} We consider the Cauchy-Riemann equations on $U$ and $V$. By the chain rule we have \begin{align} U_x&=u_x(x,-y),\\ U_y&=\frac{\partial}{\partial y}(u(x,-y))=u_y(x,-y)\times (-1)=-u_y(x,-y),\\ V_x&=-v_x(x,-y),\\ V_y&=\frac{\partial}{\partial y}(-v(x,-y))=-v_y(x,-y)\times (-1)=v_y(x,-y), \end{align} where $\times (-1)$ comes from $\frac{\partial(-y)}{\partial y}=-1$. Therefore we have $$ U_x=V_y,\quad U_y=-V_x, $$ since $u, v$ satisfy $u_x=v_y,\, u_y=-v_x.$

$E$ be a domain, define $E^*=\{z \in C: \overline{z}\in E\}$ $f: E \to C$ is analytic, then $f^*(z)=\overline{f(\overline{z})}$ analytic on $E*$.

1 Answers1

$E$ be a domain, define $E^=\{z \in C: \overline{z}\in E\}$ $f: E \to C$ is analytic, then $f^(z)=\overline{f(\overline{z})}$ analytic on $E*$.