edited alternate proof

1 files changed, 4 insertions, 4 deletions

diff --git a/ClassJuly5.page b/ClassJuly5.page
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+++ b/ClassJuly5.page
@@ -21,10 +21,10 @@ So, if you only care about holomorphic functions you don't need to worry about t
 You may find it helpful to think about other ways of deriving Theorems 1 and 2.  For an alternate proof of Theorem 1 (which may be more comprehensible, since it doesn't involve any confusing changes of coordinates), see Problems 7 and 8 of Problem Set 3.
 
 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 $f$ 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 $c_{-1} = 0$.
+$$c_{-1} = \frac{1}{2 \pi i}\int_{\gamma_r} f(z) dz$$
+where $\gamma$ is a small circle of radius $r$ that goes counterclockwise around the origin.  Taking the limit as $r \to 0$, we find ourselves integrating over a circle of radius $0$.  Therefore,  
+$$ c_{-1} = \frac{1}{2 \pi i} \lim_{r \to 0} \int_{\gamma_r} f(z) dz = \frac{1}{2 \pi i} \int_{\gamma_0} f(z) dz = 0  $$
+and we conclude that $c_{-1} = 0$.
 
 Now let's prove that $c_{-2}$ has to be zero.  Consider the function 
 $$ g(z) = z f(z) = \cdots \frac{c_{-3}}{z^2} + \frac{c_{-2}}{z} + c_{-1} + c_0 z + c_1 z^2 + \cdots $$