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Suppose that $f:\mathbb C \to \mathbb C$ is an analytic function. Then $f$ is a polynomial if (select all that apply):

(A) For any point $a\in \mathbb C$, if $f(z)=\sum_{n=0}^\infty a_n(z-a)^n$ is a power series expansion at $a$, Then $a_n=0$ for atleast one $n$.

(B) $\lim _{\left|z\right|\to \infty }\left|f\left(z\right)\right|=M$

(C) $\lim _{\left|z\right|\to \infty }\left|f\left(z\right)\right|=\infty $

(D) $\left|f\left(z\right)\right|\le M\left|z\right|^{n\ }$ for $|z|$ sufficiently large and for some $n$.

I know that (B) is false and (C) is true. I can prove it. (D) is also true. How do I prove (A) is true? Given in the Answer key that (A) is true.

For any point $a\in \mathbb C$, if $f(z)=\sum_{n=0}^\infty a_n(z-z_o)^n$ is a power series expansion at $a$, Then $a_n=0$ for at least one $n$.

Consider the power series with $a_{2n+1}=0$ and $a_{2n}=1$ this is not a polynomial. How do (A) is true?

vidyarthi
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    How did you prove (b) is false? –  mheldman Jun 12 '19 at 02:02
  • Note that (A) says that for any point the series expansion has a vanishing term. You have exhibited an example for which that happens in one point, so it does not follow that the associated series is a polynomial. – qualcuno Jun 12 '19 at 02:05
  • https://math.stackexchange.com/questions/1361377/lim-limits-z-to-infty-fz-infty-show-that-f-is-a-polynomial @mheldman –  Jun 12 '19 at 02:15
  • @Unknownx well, that question says that if the limit is infinity along every path then it is a polynomial. This is not the same as saying that if the limit of the magnitude exists then it is not a polynomial. Remember if (B) holds then $f$ is a bounded, analytic function –  mheldman Jun 12 '19 at 02:17

2 Answers2

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For A, Consider $$B:=\{z\in \Bbb C: f^{(n)}(z)=0\;\text{for some}\; n \in \Bbb N \}=\bigcup_n\{z\in \Bbb C: f^{(n)}(z)=0\}$$

Here $B$ is uncountable. That means, atleast one set in the union is uncountable. Thus, $\exists k$ so that $\{z: f^{(k)}(z)=0\}$ is uncountable, so it has limit point in $\Bbb C$ and hence result follows!

Chinnapparaj R
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$B$ is true, by Liouville's theorem.

$A$ is sufficient (see the other answer).

$C$ is apparently true (see the link in the comments).

For $D$ it means $\mid f(z)\mid=O(\mid z^n\mid)$. It's sufficient. See this.

  • https://math.stackexchange.com/questions/1361377/lim-limits-z-to-infty-fz-infty-show-that-f-is-a-polynomial –  Jun 12 '19 at 02:15
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    (A) He wants the series expansion at any point having at least one zero coefficient. It means that given $z_{0}$, $f^{n}\left(z_{0}\right)=0$ for at least one $n$. $\cos z$ doesn't have this property.

    For (C), note that $\lim_{x\rightarrow-\infty}e^{x}=0$, so I can't see how $e^{z}$ diverges in all directions in the complex plane.

     – eranreches Jun 12 '19 at 02:16
  • Ok. I blew it. @Unknownx –  Jun 12 '19 at 02:20