The problem is $\int z^n (1-z^m)$ along the circle contour parameterised by $2e^{it}$, $t = 0$ to $2\pi$, where $n, m < 0$.
I've used the keyhole contour argument to show that the radius of the circle doesn't matter as long as it's no less than 2, and the notes say this is useful. We can effectively set our own value $R > 2$ for the radius, so the value of the integral is independent of the radius of the contour. I don't know how to proceed from here.
We were told not to use the residue theorem.
I will use $m,n \ge 1$ instead.
Let $\gamma(t) = 2 e^{it}$, then $I = \int_\gamma ( {1 \over z^n} - {1 \over z^{m+n}} ) dz $.
If $k=1$, then $\int_\gamma {1 \over z} dz = \int_0^{2 \pi} {1 \over 2e^{it}}i 2e^{it} dt = 2 \pi i$.
If $k >1$, then $\int_\gamma {1 \over z^k} dz = \int_0^{2 \pi} {1 \over 2^ke^{ikt}}i 2e^{it} dt = 0 $.
Hence, if $n=1$, we have $I = 2 \pi i$, if $n >1$ then $I = 0$.