Suppose $f$ is entire and that in every power series $f(z)=\sum_{n=0}^\infty c_n(z-a)^n$ at least one coefficient is $0$. Prove that $f$ is a polynomial.

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    – Did
    Aug 16, 2013 at 0:22

1 Answer 1


Hint: For each complex number $a$, there exists an $n$ for which $c_n(a) = 0$, where $c_n(a)$ is the $n$th coefficient for the expansion at $a$. There are uncountably many complex numbers, but only countably many naturals, so....

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    $\begingroup$ Ah so there exists a $n$ with uncountable number of zeros. Such that $0=c_n(a)=\frac{f^{(n)}(a)}{n!}$ for uncountably many $a$. Since each derivative is entire, there are only countable number of zeros. Hence the $n$th derivative is 0, and $f$ is a polynomial. $\endgroup$
    – anon
    Aug 16, 2013 at 0:59
  • $\begingroup$ @user61527:I have one doubt regarding above question.Is it necessary that n is fixed?If n is not fixed then how can we say that $c_n$ is a function of $z$? $\endgroup$ May 19, 2018 at 6:29
  • $\begingroup$ @ASHWINISANKHE Recall from the expansion of power series that $c_n(a)=\frac{f^{(n)}(a)}{n!}$ $\endgroup$
    – Brozovic
    Jul 11, 2020 at 14:13

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