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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.

Hurjui Ionut
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  • Are $a,b$ positive integers? – zhw. Jul 03 '18 at 14:11
  • Yes, of course! – Hurjui Ionut Jul 03 '18 at 14:13
  • Try to show that the product of two open maps is open. I think this is true, however I could be wrong. – Sean Haight Jul 03 '18 at 14:33
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    @SeanHaight $z,\bar z$ are open, but their product isn't. This explains the $a\ne b$ condition. – zhw. Jul 03 '18 at 14:40
  • You need to assume $\Omega$ is connected. – zhw. Jul 03 '18 at 14:47
  • Yes, @zhw, you are so right $\Omega$ needs to be connected. – Hurjui Ionut Jul 03 '18 at 14:53
  • Does neighborhood really mean "open neighborhood"? I would not expect that $h(U)$ is open for any open neighborhood of $0$ in $\Omega$, but only for sufficiently small $U$. In that iInterpretation $h(U)$ would be a neighborhood of $0$ for any neighborhood $U$ of $0$ in $\Omega$ and it would be irrelevant whether $\Omega$ is connected. – Paul Frost Jul 03 '18 at 14:59
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    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 $f(U).$ – Hurjui Ionut Jul 03 '18 at 15:11
  • Perhaps it is useful to observe that $u(z)= z^a \overline{z}^b$ maps $B(0,r)$ to $B(0,r^{a+b})$. Let us assume that $a > b$, the case $a < b$ is similar. We have $\lvert u(z) \rvert = \lvert z \rvert^{a+b}$ and $u(z) = \lvert z \rvert^{2b} z^{a-b}$. For $w = se^{it}$ let $w' = s^{1/(a+b)} e^{it/(a-b)}$. Then $u(w') = w$. – Paul Frost Jul 03 '18 at 15:57
  • @PaulFrost So $u$ is open map. How is this helpful? – Hurjui Ionut Jul 03 '18 at 16:15
  • I don't know whether it helps, it was just an observation. If $f,g$ are constant, we are ready. – Paul Frost Jul 03 '18 at 17:25

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