When dealing with a contour along a unit circle, we can set $|z| = 1$ and $z(t) = e^{it}$ so that $\frac{d z}{dt} = ie^{it}$ with $t\in [0,2\pi]$.
Find the integral of:
$$\int_{\Gamma}(z^7+z^4) dz$$
My own attempt.
$$\int_{\Gamma} (z^7+z^5)dz = \int_0^{2\pi}((e^{it})^7+(e^{it})^5)(ie^{it})dt $$ $$i\int_0^{2\pi}e^{8it}+e^{6it}dt$$
Am I doing it right? It doesn't seem like the right answer but I can't find anything that says it's not. How would it be done using the Residue Theorem?
No, it is not correct. It should be\begin{align}\int_\Gamma z^7+z^5\,\mathrm dz&=\int_0^{2\pi}\left(\left(e^{it}\right)^7+\left(e^{it}\right)^5\right)ie^{it}\,\mathrm dt\\&=i\int_0^{2\pi}e^{8it}+e^{6it}\,\mathrm dt\\&=i\int_0^{2\pi}\cos(8t)+i\sin(8t)+\cos(6t)+i\sin(6t)\,\mathrm dt\\&=i\left[\frac{\sin(8t)}8+i\frac{-\cos(8t)}8+\frac{\sin(6t)}6+i\frac{-\cos(6t)}6\right]_{t=0}^{t=2\pi}\\&=0.\end{align}