I have $z(t)=t(1-t)e^t + \cos(2 \pi \cdot t^3)i$ with $0 \le t \le 1$ and need to evaluate the line integral of $e^{z^2}$.
I know that the endpoints are $z(0)=z(1)=0+i$, so the line is a closed curve. I also know that $e^{z^2}$ is an analytic function, so we can use Cauchy's integral formula here.
I have a decent understanding of Cauchy's, but still kind of confused. Doesn't it make us "pick" a point that lies inside the curve (since the denominator is $(z-a)$). How am I supposed to pick a point within this curve for this problem? It feels fairly arbitrary. Thanks!
Cauchy's Integral Theorem says that if $f$ is analytic on a simply connected domain $D$ and $\gamma$ is a closed, rectifiable curve in $D$ then
$$ \oint_\gamma f = 0. $$
Cauchy's Integral Formula says that under the same hypotheses and if $\pmb\gamma$ winds around $\pmb a$ once counterclockwise, then
$$ \frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z - a} dz = f(a). $$
The point being that there are two theorems which are related but do different things for you.
One could also say $$ \frac{n!}{2\pi i}\oint_\gamma \frac{f(z)}{(z - a)^{n+1}} dz = f^{(n)}(a). $$ The Integral Theorem is with $n = -1$ (almost; you need to get rid of the $n!$ in this case) and the Integral Formula is with $n = 0$.