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I have to evaluate the following integral: $ \int_{\gamma} \frac{1}{z^3-z^2+z-1} dz$, $ \gamma: [0, 2\pi] \rightarrow \mathbb{C}, \gamma(t)=1+e^{2it} $

My idea was to split the integrand into partial fractions and apply cauchys integral formula. Then i get:

$\int_{\gamma} \frac{1}{z^3-z^2+z-1} dz = \frac{1}{2}\int_{\gamma} \frac{1}{z-1}dz- \frac{1}{4} \int_{\gamma} \frac{1-i}{z-i}dz- \frac{1}{4} \int_{\gamma} \frac{1-i}{z+i}dz $ The last 2 integrals are 0, because $ \pm i $ arent enclosed by the given contour. For the first one, i get using the cauchy integral formula: $ 2 \pi i \cdot ind_{\Gamma} \cdot 2 = \int... $. Therefore $\int... = 2 \pi i \cdot 2 \cdot 2 = 8 \pi i$ Is that right?

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  • $\begingroup$ Is my solution right? $\endgroup$ – Sarah34 Nov 25 '18 at 9:49

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