This is the problem:

Let $0\in\Omega\subset \mathbb{C}$ connected, open neighborhood of the origin. Let $f,g:\Omega\to\mathbb{C}$ holomorphic functions such that $f(0)\neq0\neq g(0).$ Let $h:\Omega\to\mathbb{C}$ to be $h(z)=f(z)\overline{g(z)}z^a\overline{z}^b$ with $a\neq b,$ positive integers. Take $U$ a neiberhood of the origin. show that $h(U)$ is a neighborhood of the origin.

Clearly, the only thing I have to prove is that $h(U)$ is open. $h$ is not holomorphic but I've tried to mimic the proof for the open map theorem with no succes. I've tried to some arguments with isolated zeros of $f$ and $g$ but no luck.

  • $\begingroup$ Are $a,b$ positive integers? $\endgroup$ – zhw. Jul 3 '18 at 14:11
  • $\begingroup$ Yes, of course! $\endgroup$ – Hurjui Ionut Jul 3 '18 at 14:13
  • $\begingroup$ Try to show that the product of two open maps is open. I think this is true, however I could be wrong. $\endgroup$ – Sean Haight Jul 3 '18 at 14:33
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    $\begingroup$ @SeanHaight $z,\bar z$ are open, but their product isn't. This explains the $a\ne b$ condition. $\endgroup$ – zhw. Jul 3 '18 at 14:40
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    $\begingroup$ Yes, you are right! i just have to show that i can find a radius $r>0$ such that the $B(0,r)$ (ball centred at origin with radius $r$) is contained in $f(U).$ $\endgroup$ – Hurjui Ionut Jul 3 '18 at 15:11

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