I'm trying to practice by computing
$$\int_{\gamma(0,1)} \frac{5z^²-8}{z^3-2z^2}dz$$
I've first tried to expand in partial fractions and then set $z=e^{i\theta}$, but while it first simplifies quite nicely, it then ends up presenting some integrals that aren't really nice (I get one that's not converging).
I'm wondering whether maybe there is a simpler way, or if I've just made a mistake in my computation and could see a right one... Is the integrand actually differentiable on all open regions of $\mathbb{C}$ not containing $0$ and $2$? I can't prove it but I feel it's true, then I could apply the Fundamental Theorem of Calculus for path integrals...
Can you help me clarify all this?
Thank you very much!
You don't even need to use partial fractions: $$ \int_{\gamma(0,1)} \frac{5z^²-8}{z^3-2z^2}dz = 5\int_{\gamma(0,1)}\frac{1}{z-2}dz - 8\int_{\gamma(0,1)} \frac{1}{z^2(z-2)}dz $$ The first integral is zero since the function $\frac{1}{z-2}$ is holomorphic in a convex neighbourhood of the curve. For the second integral we use Cauchy's Theorem and again the fact that $\frac{1}{z-2}$ is holomorphic: $$ \int_{\gamma(0,1)} \frac{1}{z^2(z-2)}dz = \int_{\gamma(0,1)}\frac{\frac{1}{z-2}}{z^2}dz = \frac{2\pi i}{1!}f^{(1)}(0) = 2\pi i\frac{-1}{(z-2)^2}\left.\right|_{z=0} = -\frac{\pi i}{2} $$ Which yields $I = 4\pi i$.