This is the integrand of a complex integral:
$$\frac{o(\zeta - z)}{\zeta - z}$$
The ensuing discussion says that this can be made as small as desired [by confining $\zeta$ close to $z$].
In general I thought little-$o$ notation implied that given two functions,
$f(x) = o(g(x))$ as $x \rightarrow a$ if $$\lim_{x \to a}\frac{f(x)}{g(x)} = 0$$
I would appreciate help in seeing how to apply this to the above integrand to see how it can be made as small as desired.
Thanks
If we take your definition of little-O, taking $g(\zeta) = \zeta -z$, taking $f$ to be the little-o function of $g$ in the numerator, and $D$ to be the domain you have:
$$\lim_{\zeta \to z}\frac{f(\zeta)}{g(\zeta)} = 0$$ implies
$$\forall \epsilon>0, \exists \delta>0: \forall \zeta \in D: 0<|\zeta - z| < \delta \implies |\frac{f(\zeta)}{g(\zeta)}|<\epsilon$$
$$\implies |\frac{f(\zeta)}{\zeta-z}|<\epsilon$$
By the definition of limits.
So your however small ($\epsilon$) you want the magnitude of your quotient ($|\frac{f(\zeta)}{\zeta-z}|$) to be, you can bound it below that just by finding the appropriate $\delta$.
So your integrand can be as small as you wish. If you're asking about the integral itself, look up the estimation lemma.
http://en.wikipedia.org/wiki/Estimation_lemma