Proving complex integrals Hi I'm a bit lost with the following question and was wondering if anyone could help:
Prove that $\ \int_c (z-a)^n dz = 2\pi i $ iF c is given by $\ z(t) = a+ e^{it}, 0 \le t \le 2 \pi $ and $\ n = -1 $
Show that the integral is zero for all other values of n ∈ Z.
any help would be greatly appreciated!

Just compute the integral:
$$
\int_c (z-a)^n \, dz = \int_0^{2\pi} \! e^{nit} i e^{it} \, dt
= \int_0^{2\pi} \! i \, e^{(n+1)it} \, dt
$$
When $n=-1$, the integrand reduces to $i$ and the integral is $2\pi i$.
When $n\ne -1$, the integral is zero by the fundamental theorem of calculus, because $e^{(n+1)it}$ is periodic with period $2\pi$.