The question is :Compute the following line integral:
$\int_\gamma(z^2+3z +4) dz$,
where $\gamma$ is the circle ${|z|=2}$,oriented counterclockwise.
Here is my solution:
f(z)=$(z^2+3z +4)$,
$\gamma(t)=2e^{it}$ ,$0 \leq t \leq 2\pi$
${\gamma(t)}'$=$2ie^{it}$
$f(\gamma(t))=4e^{2it}+6e^{it}+4$
$\int_\gamma f(z) dz$=$\int_{0}^{2\pi} f(\gamma(t))\gamma(t)'dt$
=$\int_{0}^{2\pi} (4e^{2it}+6e^{it}+4)2ie^{it}dt$
=2i$\int_{0}^{2\pi} (4e^{3it}+6e^{2it}+4e^{it})dt$
=$2i(\frac{4}{3i}e^{3it}+\frac{6}{2i}e^{2it}+4e^{it})|_{0}^{2\pi}$
=0
Since $e^{it}|_{0}^{2\pi}$=$cos(2\pi)+isin(2\pi)-cos(0)-isin(0)$
=0
Similarly, $e^{3it}|_{0}^{2\pi}$= $e^{2it}|_{0}^{2\pi}$=0
