Integrating $\frac{1}{z^4}$ over the unit circle I need to integrate
$$\int_{\vert z \vert = 1} \frac{1}{z^4}dz$$
I appeal to
$$\int_\gamma f(z) dz = \int_a^b f(\gamma(t))d\gamma(t)$$
And I get
$$\int_0^{2 \pi} \frac{1}{e^{4 i t}} ie^{it}dt$$
this equals
$$i\int_0^{2 \pi}e^{-3 i t} dt=-\frac{1}{3}(e^{-3 i t})\bigg\vert_0^{2 \pi}=0$$
Did I do this correctly?

That's perfect. Another way to solve it that can be useful for complicated integrals is the residue theorem. Since the only pole of the function inside your contour is at $z=0$, and $Res(f,0)=0$, we get that
$$\int_\gamma \frac{1}{z^4}dz=2\pi i Res(f,0)=0$$
