complex integral, partial fractions 2

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?

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