Can anyone see a way to do the integral under the curve $\gamma(t)=2e^{it}, t\in[0, 2\pi]$:
$$\int _\gamma \frac{2z}{z^2-2}dz$$
My intention is to use the definition (This exercise appears right after the definition of complex integral). After some accounts I arrived at:
$$\int _\gamma \frac{2z}{z^2-2}dz= \int_o^{2\pi}\frac{i(cos(2t)+isen(2t))}{2(cos(2t)+isen(2t))+1}dt=0+4i\int_0^{2\pi}\frac{2+cos(2t)}{5+4cos(2t)}dt$$
But calculating this last integral is being very difficult. Does anyone see an easier way to solve this?
Notice that you have two singularities $$ \int_\gamma \frac{2z}{z^2 -2}dz = \int_\gamma \frac{2z}{(z+\sqrt{2})(z-\sqrt{2})} dz $$ You can make an argument using symmetry or calculating by using the Residue theorem. That is, $$ \lim\limits_{z\to \pm \sqrt{2}} \frac{2z}{z \pm \sqrt{2}} = \pm 1 $$