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Find the most general function $f: \mathbb{C} \rightarrow \mathbb{C}$ such that $f$ is entire and $\exists C > 0$ with $|f(z)| \leq C|z|^2$ $\forall z \in \mathbb{C}$.

I'm really not sure where to start with this problem.

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  • $\begingroup$ Have a look at this example and apply the same strategy. $\endgroup$ – rtybase Mar 19 '19 at 20:33
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Hint: Use the Cauchy integral formula on disks $D_R(0)$ and send $R \to \infty$ (this is fine since $f$ is entire). Doing so, you should be able to show that, above some order, the derivatives of $f$ vanish.

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  • $\begingroup$ We have not learned how to use the Cauchy integral formula on disks. Is there another way to do it? $\endgroup$ – Amanda Lococo Mar 19 '19 at 19:15
  • $\begingroup$ What tools do you have at your disposal? $\endgroup$ – Gary Moon Mar 19 '19 at 19:16
  • $\begingroup$ Cauchy integral test for complex functions in one dimension, Weierstrauss M-test, Lipschitz functions, uniform convergence, ratio test, domain of convergence, points of singularity. That's all we have covered so far in the course $\endgroup$ – Amanda Lococo Mar 19 '19 at 19:19
  • $\begingroup$ Have you talked about power series for holomorphic functions? If so, since $f$ is entire, it is everywhere equal to its power series centered at $z=0$. You could write $f(z) = \sum_{n=0}^\infty a_n z^n$ and go from there. $\endgroup$ – Gary Moon Mar 19 '19 at 19:22
  • $\begingroup$ We are talking about that next, but is not something we covered for this current exam, which could contain this question. The only other consideration is perhaps this practice exam was made with the intent we would have earned holomorphisms by now? $\endgroup$ – Amanda Lococo Mar 19 '19 at 19:27
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Hint: Observe that $f(z)/z^2$ is a bounded, entire function (can you see why it's holomorphic at $0$?)

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  • $\begingroup$ I see why it is holomorphic at 0, but still don't understand how to find the most general function that satisfies the equation. Can I have some more guidance? $\endgroup$ – Amanda Lococo Mar 19 '19 at 20:25
  • $\begingroup$ My intended solution proceeded by applying the Liouville's theorem to $f(z)/z^2$, but from your comments under the other answer it appears you have not covered that yet. Indeed, it seems that you have covered basically nothing regarding holomorphic functions, so I must say I have no idea how you could approach this question. $\endgroup$ – Wojowu Mar 19 '19 at 20:44
  • $\begingroup$ I was mistaken underneath. We have been using the term analytic, not holomorphic, and I did not realize it was the same since it has never been introduced to me that way $\endgroup$ – Amanda Lococo Mar 19 '19 at 20:50
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As $f(z)$ is entire, it can be expressed as a Taylor's series $$f(z)=\sum_{n=0}^\infty a_nz^n$$therefore the most general form comes from $a_n=0$ for $n>2$ and $a_0=a_1=0$ so we have $$f(z)=a_2z^2$$

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