how to find these contour integrals?

• Feb 20th 2009, 01:09 PM
szpengchao
how to find these contour integrals?
$\displaystyle \int_{C}{z \ \ bar \ \ dz}$ , C is along the circle |z|=1
and

another case, C along the circle |z-1|=1.

$\displaystyle \int_{C}{\sqrt{z} dz}$. C is |z|=1, and |z-1|=1

i think the first part is the same for both cases which is 2pi*i.

and the second part is 0 for both cases by Cauchy Integral Theorem.

But i am not sure if that can be applied. Because sqrt z is only analytic on C\ { x axis } . Therefore, out circles are not connected?
• Feb 20th 2009, 05:32 PM
HallsofIvy
No deep theorems required here. Just parameterize the curve and integrate.

On the circle |z|= 1, $\displaystyle z= e^{i\theta}$ so $\displaystyle z*= e^{-i\theta}$ and $\displaystyle dz= ie^{i\theta}d\theta$. Of course to cover the entire circle $\displaystyle \theta$ must go from 0 to $\displaystyle 2\pi$. The integral is $\displaystyle \int_0^{2\pi}(e^{-i\theta})(ie^{i\theta}d\theta= i\int_0^{2\pi}d\theta$

On the circle |z-1|= 1, $\displaystyle z= 1+ e^{i\theta}$ so $\displaystyle z*= 1+ e^{-i\theta}$ and $\displaystyle dz= -ie^{-i\theta}$.

If $\displaystyle z= e^{i\theta}$ $\displaystyle \sqrt{z}= z^{1/2}= e^{i\theta/2}$.

On $\displaystyle z= 1+ e^{i\theta}$ $\displaystyle \sqrt{z}= \sqrt{1+ e^{i\theta}}$.
• Feb 20th 2009, 10:19 PM
szpengchao
yeah
yeah. i did those steps too. but then i found the last case is a bit difficult to calculate.