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| -rw-r--r-- | ClassJuly5.page | 2 |
1 files changed, 1 insertions, 1 deletions
diff --git a/ClassJuly5.page b/ClassJuly5.page index d740cf7..8f59ebf 100644 --- a/ClassJuly5.page +++ b/ClassJuly5.page @@ -27,7 +27,7 @@ $$c_{-1} = \frac{1}{2 \pi i}\int_{\gamma_r} f(z) dz$$ where $\gamma$ is a small circle of radius $r$ that takes one counterclockwise turn around the origin. Taking the limit as $r \to 0$, we find ourselves integrating over a circle of radius $0$: $$ c_{-1} = \frac{1}{2 \pi i} \lim_{r \to 0} \int_{\gamma_r} f(z) dz = \frac{1}{2 \pi i} \int_{\gamma_0} f(z) dz = \frac{1}{2 \pi i} \int_0^{2\pi} f(0) \frac{d\gamma_0}{dt} dt$$ But $\gamma_0(t) = 0$, hence $\frac{d \gamma_0}{dt} = 0$. We conclude that -$$c_{-1} = \frac{1}{2 \pi i} \int_0^{2\pi} 0 dt = 0 $. +$$c_{-1} = \frac{1}{2 \pi i} \int_0^{2\pi} 0 dt = 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 $$ |