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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 c1793ba..d740cf7 100644 --- a/ClassJuly5.page +++ b/ClassJuly5.page @@ -26,7 +26,7 @@ First let's prove that $c_{-1}$ is zero. Since $c_{-1}$ is the residue of $f$ a $$c_{-1} = \frac{1}{2 \pi i}\int_{\gamma_r} f(z) dz$$ 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 $\gamma_0(t)$ is constant, hence $\frac{d \gamma_0}{dt} = 0$. Therefore, +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 $. Now let's prove that $c_{-2}$ has to be zero. Consider the function |