From 51e542f89644b3cdd564f9caa0780481690d3520 Mon Sep 17 00:00:00 2001 From: luccul Date: Tue, 13 Jul 2010 14:08:40 +0000 Subject: grammar --- ClassJuly5.page | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/ClassJuly5.page b/ClassJuly5.page index 84173e9..4282c4a 100644 --- a/ClassJuly5.page +++ b/ClassJuly5.page @@ -22,7 +22,7 @@ You may find it helpful to think about other ways of deriving Theorems 1 and 2. An alternate proof of Theorem 2 goes as follows: Since $f$ is holomorphic on a disk, it has a Laurent expansion. The statement of Theorem 2 says that the negative terms in this Laurent expansion are zero. First let's prove that $c_{-1}$ is zero. Since $c_{-1}$ is the residue of $f$ at zero, it is given by $$c_{-1} = \int_{\gamma_r} f(z) dz$$ -where $\gamma$ is a small circle of radius $r$ that goes counterclockwise around the origin. As we shrink the radius of this circle, its length goes to zero. On the other hand since $f(z)$ tends to $f(0)$. Taking the limit as $r \to 0$, +where $\gamma$ is a small circle of radius $r$ that goes counterclockwise around the origin. As we shrink the radius of this circle, its length goes to zero. On the other hand since $f$ is holomorphic, $f(z)$ tends to $f(0)$ as $z$ tends to zero. Taking the limit as $r \to 0$, $$ c_{-1} = \lim_{r \to 0} \int_{\gamma_r} f(z) dz = \lim_{r \to 0} 2 \pi r f(0) = 0 $$ so we conclude that $c_{-1} = 0$. -- cgit v1.2.3