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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.

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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
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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 $f$).

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$k$ exists since $f$ is not constant, of course... –  Thomas Andrews Sep 5 '12 at 17:02
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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.

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