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This question was asked in a masters exam for which I am preparing and I was unable to think about it. So, I am asking it here.

Question: Let f be a continuous function from $\mathbb{C} \to \mathbb{C} $ such that $f(z^2+2z-6) $ is an entire function. Show that f is an entire function.

As fog is entire where g is $ z^2+2z-6$ . So,$ (f'og )\times g' $ exists for all but how can I use it to verify analyticity of f?

Please help.

Thanks!!

1 Answers1

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Hint: Let $g(z)=z^2+2z-6$. Now, $g'(z)\not=0$ for all $z\not=-1$. So, for each $z\not=-1$ we can find two small open sets $U_z$ and $V_z$ such that $g\big|U_z\to V_z$ is bi-holomorphic. So, considering $\big(f\circ g\big)\circ \big(g\big|U_z\to V_z\big)^{-1}$ we can say $f$ is holomorphic on $\Bbb C\backslash g^{-1}(-1)$. Now, $f$ is continuous on $\Bbb C$, so by Riemann removable singularity theorem $f$ is holomorphic. Note that $g^{-1}(-1)$ can have at most two elements.

Sumanta
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  • Can you please tell which result you used for the assertion " for each z≠−1we can find two small open sets$ U_z $ … is biholomorphic"? –  Nov 30 '20 at 17:29
  • Do you mind replying to my question asked in comments? –  Jan 29 '21 at 09:57
  • Oh sorry, I have forgotten. I will write. – Sumanta Jan 29 '21 at 10:31
  • First, note that $g$ is holomorphic, and for any $z\not=-1$ we have $g'(z)\not=0$. Now, using this link for each $z\not=-1$ find a nbd $U_z$ of $z$ so that $g\big|U_z$ is injective. Since, $g$ is non-constant holomorphic $g\big|U_z$ is an open map. So, $V_z:=g(U_z)$ is an open set. – Sumanta Jan 29 '21 at 10:45
  • Hence, $g|U_z\to V_z$ is bi-holomorphic by this. You may assume, $U_z$ and $V_z$ are connected by shrinking them, if necessary, using continuity. – Sumanta Jan 29 '21 at 10:46
  • Let me know if you have any question(s). – Sumanta Jan 29 '21 at 10:48