1. ## so confused

Using the parametric representation
z(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.

2. Originally Posted by sandy
Using the parametric representation
z(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.

I'm not quite sure what you're asking. Is it to evaluated $\displaystyle \int_C \frac{dz}{z}$ with $\displaystyle C=\left\{e^{it}:t\in[0,\pi]\right\}$?

3. 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 $\pi$, 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.

4. Originally Posted by Drexel28
I'm not quite sure what you're asking. Is it to evaluated $\displaystyle \int_C \frac{dz}{z}$ with $\displaystyle C=\left\{e^{it}:t\in[0,\pi]\right\}$?
Yes thats reight im asking that and the integral is from -1 to 1

5. Originally Posted by HallsofIvy
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 $\pi$, 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.