Cauchy-Riemann equations and holomorphic functions

Question

Suppose $f(z) = u(x,y) + iv(x,y)$ is holomorphic. Prove that if $f(\overline{z})$ is holomorphic as well, then $f$ is a constant function. I'm having trouble showing that the partial derivatives $u_x, u_y, v_x, v_y$ are all zeros. I tried to list out the Cauchy-Riemann equations for $f(z)$ and $f(\overline{z})$ but can't figure out how to proceed to show the derivatives are equal to zeros. So far what I have is that $$ \begin{align*} u_x(x,-y) &= v_y(x,-y) \\ u_y(x,-y) &= -v_x(x,-y). \end{align*} $$ But using Cauchy-Riemann on $f(z)$ gives $$ u_x(x,y) = v_y(x,y) \quad \quad u_y(x,y) = -v_x(x,y), $$ which doesn't really give me much information about the derivatives. Could someone help me with this?

Answer 1

Actually, applying the Cauchy-Riemann equations to $f(\overline z)$ tells you that you have, due to the chain rule,$$u_x(x,-y)=-v_y(x,-y)\quad\text{and}\quad-u_y(x,-y)=-v_x(x,-y)$$and the second equality is equivalent to $u_y(x,-y)=v_x(x,-y)$. Since this occurs for each $x$ and each $y$, you have$$u_x(x,y)=-v_y(x,y)\quad\text{and}\quad u_y(x,y)=v_x(x,y).\tag1$$But you also have (applying the Cauchy-Riemann equations to $f$) that$$u_x(x,y)=v_y(x,y)\quad\text{and}\quad u_y(x,y)-v_x(x,y).\tag2$$And it follows from $(1)$ and $(2)$ that each of the functions $u_x$, $u_y$, $v_x$, and $v_y$ is the null function. Therefore, $f$ is constant.