Let $f$ be a holomorphic function on open set $A$ such that $(Im(f(z))^3 + (Re(f(z))^4 =5.$ Could anyone advise me on how to use Open mapping theorem to prove $f$ is constant? Hints will suffice. Thank you.

  • $\begingroup$ You need to assume $A$ is both open and connected. Otherwise, if you have two disconnected components of $A$, you can define $f(z) = 5^{1/4}$ in the first component, and $f(z) = i5^{1/3}$ in the second. $\endgroup$ – Michael Sep 4 '14 at 5:45
  • $\begingroup$ You might determine if the image of $f$ is open or not. $\endgroup$ – Michael Sep 4 '14 at 5:46
  • $\begingroup$ Noted. So we need to show $\{(x,y): x^4 + y^3 = 5\}$ is not open? $\endgroup$ – Alexy Vincenzo Sep 4 '14 at 5:52
  • $\begingroup$ If you can show that, then... $\endgroup$ – Michael Sep 4 '14 at 5:53
  • $\begingroup$ I am still trying to show... $\endgroup$ – Alexy Vincenzo Sep 4 '14 at 5:59

Assume $f$ is not constant on the domain $\Omega\subset{\mathbb C}$. Then there is a point $z_0\in \Omega$ with $f'(z_0)\ne0$, and $f$ maps a suitable neighborhood $U$ of $z_0$ bijectively onto a full neighborhood $V$ of the point $f(z_0)=:u_0+iv_0$.

By assumption the point $(u_0,v_0)$ lies on the set $$\gamma:=\{u+iv\>|\>u^4+v^3=5\}\ .$$ The set $\gamma$ is a level set of the function $\phi(u,v):=u^4+v^3$. Since $\nabla\phi(u,v)=(4u^3,3v^2)$ is $\ne(0,0)$ for all points $(u,v)\in\gamma$, the implicit function theorem tells us that there is a window $$W:=[u_0-h,u_0+h]\times[v_0-h,v_0+h],\qquad h>0,$$ and a function $$\psi:\quad[u_0-h,u_0+h]\to [v_0-h,v_0+h],\qquad u\mapsto v=\psi(u),$$ $\bigl($or a similar function $v\mapsto u=\chi(v)\bigr)$ such that $\psi(u_0)=v_0$ and $$\gamma\cap W=\{\bigl(u,\psi(u)\bigr)\>\bigm|\>u_0-h\leq u\leq u_0+h\bigr\}\ .$$ It follows that $V$ contains points not on $\gamma$, e.g., the points $(u_0,v_0+\delta)$ for sufficiently small $\delta>0$, contrary to assumption about $f$.

  • $\begingroup$ why so complicated? Also, it seems the asker is only asking for some hints, not a full answer. $\endgroup$ – Michael Sep 4 '14 at 13:10
  • $\begingroup$ @Michael: We all know that $\gamma$ is a curve and does not contain any open set; but I had the impression the OP wanted a proof of this. $\endgroup$ – Christian Blatter Sep 4 '14 at 13:14

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