Terris Becker edited untitled.tex  almost 8 years ago

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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}\\