Let $f$ be an analytic function such that $\operatorname{Im}(f(z))=\operatorname{Re}(f(z))^2$. Prove that $f$ is constant.

Question

How to solve this question? Also it would be great if someone could tell me from which book this question is from?

Answer 1

Real and imaginary parts of an analytic function are harmonic.

Let $u$ and $v$v be the real and imaginary parts of $f$. Then $v=u^{2}$. Show that $0=v_{xx}+v_{yy}=2(u_x^{2}+u_y^{2})$ and conclude that $u_x=u_y=0$ . Hence $u$ is a constant and so is $v$.

Answer 2

The range of $f$ has empty interior because it is a subset of $\{(x,y)\in\Bbb R^2\,:\, y=x^2\}$. Therefore $f$ is not open, which means that it's (locally) constant.

Answer 3

$\text{Re}^2(f)+\text{Im}^2(f)=|f|^2 \iff \text{Im}^2(f)+\text{Im}(f)-|f|^2=0$. As $b^2-4ac=1+4|f|^2\geq 0$ the equation has at least one non-negative solution,i.e. , $ \text{Im}(f)=c$ for some $c \geq0$. Finally $\text{Re}^2(f)=c $ only has one solution, otherwise the function $\text{Re}(f)$ would not be continuous. Therefore $f$ is constant.