Evaluate $$\int_{\gamma}\frac{z^2+2z}{z^2+4}dz$$ where the contour $\gamma$ is
1.) the circle of radius $2$ centered at $2i$, traversed once anti-clockwise.
2.) the unit circle centered at the origin, traversed once anti-clickwise.
So here we would have to use partial fractions: $$1+ \frac{2z-4}{z^2+4}.$$
Then for part 1.), $\gamma(t)=2 e^{it}+2i$.
And for part 2.), $\gamma(t)= e^{it}$.
I'm not sure what to do next to evaluate the integral for part 1.) and 2.).
Best Answer
$(1)$ Apply the residue theorem. $\int_{\gamma}\frac{z^2+2z}{(z-2i)(z+2i)}dz=2i\pi(\sum res_{z=z_k})$. Define $z_0:=z+2i.$ Thus, $res_{z=z_0}f(z)=\frac{z_k^2+2z_k}{2z_k}$, for $k=0$.
$(2)$ Notice that none of your singluar points are in your contour $\Rightarrow $$\int_{|z|=1}\frac{z^2+2z}{(z-2i)(z+2i)}dz=0$, by Cauchy Integral Formula.