I'm doing a Complex Analysis class this semester and I've got this interesting integral problem.
I have to do the following integral:
$$\int_\Gamma \frac{1}{1+z^2}dz$$
Where $\Gamma$ is the circle centered on the origin, with radius 2 (I'm talking complex plane here). I'll be referring to $\frac{1}{1+z^2} $ as $ f(z)$
My basic intuition says that the integrated function is analytic (I can show, not the purpose of this question), so according to Cauchy's Integral Theorem that the integral should be equal to $0$.
Now, the question asks the following: find the branch of $\arctan(z)$ (the antiderivative of $f(z)$), which is analytic on $\Gamma$, to evaluate the integral. I know that the principal branch has two non-analytic regions; starting in $-1$ and $1$ and extending over the axis, but I can't seem to find any branch that will turn out analytic on $\Gamma$. Any thoughts on which branch I should use?
The integrand has poles at $\pm i$, which are inside $\Gamma$. Find the residues there, and you're done.
But if you don't know the Residue Theorem, compare this integral with the integral around the concentric circle of radius $R$, where $R > 2$. Since the integrand is analytic on the region between the two circles, those integrals are equal. Now take the limit as $R \to \infty$, estimating the integral...