Let $f(z)=u+iv $ be entire function such that $v=u^2$. Is $f(z)$ constant?
My approach:
After some steps I got $|{e^{if(z)}}|=\frac{1}{e^{u^2}}.$ My question is, can I claim $\frac{1}{e^{u^2}}$ is bounded?
Please help.
2026-03-24 20:30:47.1774384247
Is the entire function with given condition constant?
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The problem is that I don't know the steps that give the equality that you mention. An alternative solution: Cauchy Riemann $u_x=v_y=(u^2)_y=2uu_y$, $u_y=-v_x=-2uu_x$. We deduce that $u_x=2u(-2uu_x)=-4u^2u_x$. So, $u_x(1+4u^2)=0$ and this implies $u$ is constant. Same argument with $v$.