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| -rw-r--r-- | ClassJuly5.page | 8 |
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diff --git a/ClassJuly5.page b/ClassJuly5.page index 4282c4a..4a012c8 100644 --- a/ClassJuly5.page +++ 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 $$ g(z) = z f(z) = \cdots \frac{c_{-3}}{z^2} + \frac{c_{-2}}{z} + c_{-1} + c_0 z + c_1 z^2 + \cdots $$ |