Show that if $f(z)$ and $\overline{f(z)}$ are analytic on domain $D$ then $f(z)=$ constant.
If I'm understanding this correctly, a complex function and its conjugate should just be reflecting functions on the complex space on the y axis. Then $f(z)$ is constant on that one whole function. Is this the right approach? Or is there rather more elegant way to prove it?
Not sure if it is a more elegant way (you Will be the judge), but it can be done by the Cauchy-Riemann equation.
Since $f(z)=u(x, y)+iv(x, y)$ is analytic, $$\frac{\partial u(x, y)}{\partial x}=\frac{\partial v(x, y)}{\partial y}\tag{1}$$ Since $\overline{f(z)}=u(x, y)-iv(x, y)$ is analytic, $$\frac{\partial u(x, y)}{\partial x}=\frac{\partial (-v(x, y))}{\partial y}=-\frac{\partial v(x, y)}{\partial y}\tag{2}$$ By $(1)$ and $(2)$ $$\frac{\partial u(x, y)}{\partial x}=\frac{\partial v(x, y)}{\partial y}=0$$ $u$ is constant with respect to $x$ and $v$ is constant with respect to $y$.
Similarly, since $f(z)=u(x, y)+iv(x, y)$ is analytic, $$\frac{\partial u(x, y)}{\partial y}=-\frac{\partial v(x, y)}{\partial x}\tag{3}$$ Since $\overline{f(z)}=u(x, y)-iv(x, y)$ is analytic, $$\frac{\partial u(x, y)}{\partial y}=-\frac{\partial (-v(x, y))}{\partial x}=\frac{\partial v(x, y)}{\partial x}\tag{4}$$ By $(3)$ and $(4)$ $$\frac{\partial u(x, y)}{\partial y}=\frac{\partial v(x, y)}{\partial x}=0$$ $u$ is constant with respect to $y$ and $v$ is constant with respect to $x$.
Conclusion, $f(z)$ must be constant.
