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\begin{equation} | \begin{equation} | ||
e^{ik\pi/3}\quad\mathrm{for\ }k\in\{1,2,4,5\} | e^{ik\pi/3}\quad\mathrm{for\ }k\in\{1,2,4,5\} | ||
| − | \end{ | + | \end{equat$ion} |
| + | |||
| + | We can now calculate the integral by computing the residues and summing those on the upper half-plane (times $2i\pi$): | ||
| + | |||
| + | \begin{multline} | ||
| + | \mathrm{Res}_{k=1}\frac{1}{(z-e^{i\pi/3})(z-e^{i2\pi/3})(z-e^{-i\pi/3})(z-e^{-i2\pi/3})} | ||
| + | =\\ | ||
| + | \frac{1}{(e^{i\pi/3}-e^{i2\pi/3})(e^{i\pi/3}-e^{-i\pi/3})(e^{i\pi/3}-e^{-i2\pi/3})}\,. | ||
| + | \end{multline} | ||
| + | |||
| + | The denominator is calculated easily by dealing with the geometry of | ||
| + | the unit circle (one is a sine, the other is after computing a | ||
| + | product, the translated sum of a number with its opposite). One thus | ||
| + | finds the residue: | ||
| + | |||
| + | $$1/(2i\sqrt{3}e^{i\pi/3})$$ | ||
| + | |||
| + | The other residue is found similarly as | ||
| + | |||
| + | $$1/(2i\sqrt{3}e^{-i\pi/3})$$ | ||
| + | |||
| + | so that the integral is finally, since $1/e^{-i\pi/3}+1/e^{i\pi/3}=e^{-i\pi/3}+e^{i\pi/3}=1$, simply: | ||
| + | |||
| + | $$\frac{\pi}{\sqrt3}$$ | ||
==== Question 2 ==== | ==== Question 2 ==== | ||