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Terris Becker edited untitled.tex
almost 8 years ago
Commit id: 8e8de09b52763bb8b6a08b20bb3b85fab84d7182
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\end{align*}
We may use Cauchy's Integral Formula once again because the function in the numerator is holomorphic in $\gamma_2$.
\begin{align*}
\int_{\gamma_1}\frac{dz}{(z-(\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}))(z-(-\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}))(z-(-\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}))(z-(\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}))} \int_{\gamma_1}\frac{\frac{dz}{(z-(-\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}))(z-(-\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}))(z-(\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}))}}{(z-(\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}))} &= \frac{2\pi i}{(\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}+\frac{\sqrt2}{2}-i\frac{\sqrt2}{2})(\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}+\frac{\sqrt2}{2}+i\frac{\sqrt2}{2})(\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}-\frac{\sqrt2}{2}+i\frac{\sqrt2}{2})}\\
&= \frac{2\pi i}{(\sqrt2)(\sqrt2+i\sqrt2)(i\sqrt2)}\\
&= \frac{2\pi i}{2i(\sqrt2 +i\sqrt2)}\\
&= \frac{2\pi i}{2i\sqrt2-2\sqrt2}\\