Using the parametric representationz(t) = e^(it)= cost + isint how do we evaluate the following integral
(a) integral from 1 to -1 (1/z)dz along |z|=1, the upper half plane.
If $\displaystyle z= e^{i\theta}$, then [tex]\frac{1}{z}= z^{-1}= e^{-i\theta}, of course.
As $\displaystyle \theta$ goes from 0 to [itex]\pi[/itex], z with go over the upper unit semicircle so z covers C.
$\displaystyle \int_C \frac{1}{z} dz= \int_0^\pi e^{-i\theta} d\theta$
Since it is also true that $\displaystyle e^{i\theta}= cos(\theta)+ i sin(\theta)$ you can do that integral in terms of sin and cosine.