Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm studying Silverman's complex analysis but this book seems a lot slack. I have a question in page273:

Suppose $f(z)$ is a nonconstant analytic at $z_0$ with $f(z_0)$=0. Then, by the corollary to Theorem 8.13, there exists a neighborhood of $z_0$ that contains no other zeros of $f(z)$. Thus we may express $f(z)$ as $f(z)=(z-z_0)^k F(z)$ ($k$ a positive integer), where $F(z)$ is analytic at $z_0$ with no zeros in the neighborhood or on its boundary $C$.

I can't understand the last sentence(bold fonts). How can we know there is a factor $(z-z_0)^k$ and even more $F(z)$ is analytic?

I also have Ahlfors' book so you can give me references in Silverman or Ahlfors.

share|cite|improve this question
Express $f$ as a power series at $z_0$, namely $\sum_n a_n(z-z_0)^n$. Take $k$ the smallest integer $n$ such that $a_n\neq 0$. Then you get a factor $(z-z_0)^k$ and we can defined $F(z)$. Since it's expressed as a power series, it's analytic. – Davide Giraudo Sep 5 '12 at 15:58
up vote 1 down vote accepted

Since $f$ is analytic at $z_0$, we can write $f(z)=\sum_{n=0}^{+\infty}a_n(z-z_0)^n$. Let $$k:=\inf\{n\geq 0, a_n\neq 0\}.$$ We have $$f(z)=\sum_{n=k}^{+\infty}a_n(z-z_0)^n=\sum_{j=0}^{+\infty}a_{k+j}(z-z_0)^{k+j},$$ so we define $F(z):=\sum_{j=0}^{+\infty}a_{k+j}(z-z_0)^j$. It's an analytic function, and doesn't have any zero in the neighborhood (otherwise so will have $f$).

share|cite|improve this answer
$k$ exists since $f$ is not constant, of course... – Thomas Andrews Sep 5 '12 at 17:02

In a neighborhood of $z_0$, $$ f(z)=c_0+c_1(z-z_0)+c_2(z-z_0)^2+c_3(z-z_0)^3+\dots\tag{1} $$ If $f(z_0)=0$, then $c_0 = 0$. Each subsequent term has at least one factor of $(z-z_0)$. Suppose that $c_k(z-z_0)^k$ is the first non-zero term in the series $(1)$. Divide $f(z)$ by $(z-z_0)^k$, then we get $$ F(z)=\frac{f(z)}{(z-z_0)^k}=c_k+c_{k+1}(z-z_0)+c_{k+2}(z-z_0)^2+\dots\tag{2} $$ The series in $(2)$ converges on the same neighborhood of $z_0$ that the series in $(1)$ does. Since $f(z)$ only vanishes at $z_0$ and $F(z_0)=c_k$, we get that $F(z)$ does not vanish anywhere in this neighborhood.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.