I'm trying to study for my final this Thursday and for some reason the problem that is giving me the most trouble is, according to the professor, the "easiest one". I'm just not seeing something here, it's been a long time since I've done anything in complex that looks like this. The problem is:
$$A = \left\{ \zeta: |\zeta|= \frac{1}{2}\right\}, B: = \left\{\omega : |\omega|< \frac{3}{4}\right\}. \text{ Define } f(w) : = \oint_A \frac{z(z+1)}{z^2 + 2z -w} dz.$$
I'm asked to show that $f$ is analytic in $B$ and to find $f'(0)$. I need to learn the idea behind this much more than I need an answer to this specific question, so hints or related questions are much appreciated.
Thank you fixed!
The value of the integral will sum the residues at the poles that are interior to $A$. For $\omega$ in the restricted region, the denominator has two distinct roots, so there are either $0,1$, or $2$ simple poles within $A$ and its interior.
The roots of $z^2+2z-\omega$ are $-1\pm\sqrt{1+\omega}$. Let $U=\{\omega: |\omega|<3/4\}$. Then $\sqrt{1+U}$ is two disjoint regions in $\mathbb{C}$, one with positive real part and one with negative real part. So of the two roots of $z^2+2z-\omega$, one is to the left of $-1$ and exterior to $A$.
Certainly, there are values of $\omega$ for which the integrand has a pole at $c=-1+\sqrt{1+\omega}$ within $A$. The residue at a simple pole $c$ is $$\begin{aligned} \lim_{z\to c}(z-c)\frac{z(z+1)}{z^2+2z-\omega}&=\lim_{z\to c}(z-c)\frac{z(z+1)}{(z-c)(z+\omega/c)}\\ &=\frac{c(c+1)}{c+\omega/c}\\ &=\frac{c(c^2+c)}{c^2+\omega}\\ &=\frac{c(\omega-c)}{2(\omega-c)}\\ &=\frac{(-1+\sqrt{1+\omega})(\omega+1-\sqrt{1+\omega})}{2(\omega+1-\sqrt{1+\omega})}\\ \end{aligned}$$
Define $g(\omega)$ to be this last expression.
So $f(w)=g(w)$ if $\omega$ provides one pole within $A$. If the claim is to be true, then $f(w)=g(w)$ for all $w$ in $U$. The only way that the claim could fail to be true is if for some $\omega_0\in U$, the integrand has no poles within $A$ (making $f(\omega_0)=0$). (Although this would not be a problem if $\lim{\omega\to\omega_0}g(\omega)$ equals $0$ too.)
As noted in the comments, at $\omega=0$ there are no interior poles, but that's OK since the $g(\omega)$ approaches $0$ as $\omega\to0$.
It remains to show that when $\omega\in U\setminus\{0\}$, then $|-1+\sqrt{1+\omega}|<1/2$, and therefore there will be one pole interior to $A$. Let $V=\{c: |c|<1/2\}$. We are trying to show that $k(U)\subset V$, where $k(\omega)=-1+\sqrt{1+\omega}$. Since $k$ is one-to-one on $U$ and $V$, then this is equivalent to showing $U\subset k^{-1}(V)=\{c^2+2c:|c|<1/2\}$. My argument is that the boundary curve for $U$ ($z=\frac{3}{4}e^{it})$ and the boundary curve for $k^{-1}(V)$ ($z=\frac{1}{4}e^{2it}+e^{it}$) intersect tangentially at $t=\pi$ (corresponding to $z=-3/4$) and do not intersect anywhere else. Comparing $z$-values where $t=0$ reveals that $U$'s boundary is the interior curve.