so I need to calculate the following contour integral:
$\int_{\gamma}\frac{sin(z)}{z}dz$ where $\gamma:[0,2\pi]\rightarrow\mathbb{C} $ by $\gamma(t)=e^{it}$
Now, I tried to apply the generalized Fundamental Theorem of Algebra but in order to do so I need to work with the antiderivative of $\frac{sin(z)}{z}$ and it seems overly complicated.
Instead, I want to try the "direct parameterization method":
So, first I parameterize $\gamma$: $z(t)=e^{it}$ with $0\leq t \leq 2\pi$
Then I calculate $\frac{dz}{dt}=ie^{it}$
Now, using the product rule and substituting in the integral we get:
$\int_{0}^{2\pi}\frac{sin(e^{it})}{e^{it}}ie^{it}dt$
Which, when simplified becomes:
$i\int_{0}^{2\pi}sin(e^{it})dt=0$
Does this seem right?
You have to complete the argument as follows: let $I=\int_0^{2\pi} \sin (e^{it}) dt$. Make the substitution $s=t-\pi$. We get $I=\int _{\pi}^{\pi} \sin (-e^{is})ds=-\int _{\pi}^{\pi} \sin (e^{is})ds$. For a function with period $2\pi$ the integral over $[0,2\pi]$ is same as integral over $[-\pi,\pi]$. Hence $I=-I$ which gives $I=0$.