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\end{align*}  Since the function in the numerator is holomorphic in $\gamma_1$, we can use Cauchy's Integral formula. So we have,  \begin{align*}  \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})} 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})}  \end{align*}