My next question about contour integrals is: Is it true that:
$$\int_{\delta B(0,2)} \frac{1}{1+z^2}dz = \int_{\delta B(0,2)} \frac{\frac{z}{1+z^2}}{z} dz = \left[ 2\pi i \frac{z}{1+z^2}\right]_{z=0} = 0 $$ I tried to apply cauchys integral formula but i think $\frac{z}{1+z^2}$ isn't regular in $B(0,2)$. So i tried another way to solve this: Let $\gamma : [0,2\pi] \to \mathbb{C}, \gamma(t) = 2\exp(it), \; \gamma(0) = 2,\; \gamma(2\pi) = 2$. $$\int_{\delta B(0,2)} \frac{1}{1+z^2}dz = \int_0^{2\pi} \frac{2i\exp(it)}{1+(2\exp(it))^2}dt = \dots = \left[ \arctan(2\exp(it))\right]_{t=0}^{2\pi} = 0$$ I learned earlier that this way should be prefered because of the incontinuity of $\arctan$. However wolframalpha tells me the integral is zero. So i wonder if there is another way i could try without using the continuity of $\arctan$.
Hint. Once again use the partial fraction decomposition
$$\frac{1}{1+z^2}=-\frac{\mathrm i}{2}\frac{1}{z-\mathrm i}+\frac{\mathrm i}{2}\frac{1}{z+\mathrm i}$$
thus you can rewrite one integral in a domain with two singularities into two integrals which have only one singularity in their domain. Since our most beloved integral in complex analysis yields $$\oint_{\partial B_\rho(a)}\frac{1}{z-a}=2\pi\mathrm i$$ you might quickly see why the integral over $1/(1+z^2)$ along $\partial B_2(0)$ vanishes.