Terris Becker edited untitled.tex  almost 8 years ago

Commit id: eeaeb32b223ca8d8bf42dbf00119f87fd0e110c4

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Now first look at the integral over $\gamma_1$.  \begin{align*}  \int_{\gamma_1}\frac{dz}{z^{4}+1} &= \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}))}} &= \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}))}  \end{align*}