Let $f$ be an entire function such that $$|f(z)|\leq |z| |Re(z)|$$ for all $z$. Then prove that $f$ is constant function.
I tried it as $$|f(z)|\leq |z| |Re(z)|\leq |z^2|$$ so by extended Liouville theorem $$f(z)=az^2$$ Now it remains to prove that constant $a$ is zero. $ |f^n(0)|=|\frac{n!}{2\pi}\int\frac{f(z)dz}{z^{n+1}}|\leq \frac{n!}{2\pi}\int_{|z|=R}\frac{R|Re(z)|}{R^{n+1}}|dz|.$ Now unable to proceed further. Please help. Thank you .
$f$ is zero on the imaginary axis. Using the identity theorem, it follows that $f$ is identically zero.
If you have already shown that $f(z)=az^2$ then you don't need the identity theorem: $f(iy) = 0$ for $y \in \Bbb R$ implies that $a=0$.