Compute integral of:

Integral over C of 1/z dz

where C is the unit circle centered at some point z.

|z.| >2

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- Feb 28th 2010, 02:29 PMstumped765contour integrals
Compute integral of:

Integral over C of 1/z dz

where C is the unit circle centered at some point z.

|z.| >2 - Feb 28th 2010, 07:07 PMwowsomeme
well a unit circle in 1 not z= 2

try putting z into polar form and then do a contour integral - Mar 1st 2010, 12:05 PMstumped765
the center of the circle is at a point z, |z|>2 its just saying the unit circle is shifted from the origin..

- Mar 1st 2010, 12:08 PMwowsomeme
would it be written then |z-2|>2

- Mar 1st 2010, 12:59 PMHallsofIvy
Then use something like $\displaystyle z_0$ to differentiate it from the variable $\displaystyle z$.

The**unit**circle about $\displaystyle z_0$ with radius 1 is $\displaystyle |z-z_0|= 1$ and $\displaystyle z= z_0+ e^{i\theta}$ with $\displaystyle \theta$a going from 0 to $\displaystyle 2\pi$.

Then $\displaystyle dz= ie^{i\theta}d\theta$ $\displaystyle \oint \frac{1}{z}dz= \int_0^{2\pi}\frac{ie^{i\theta}}{z_0+ e^{i\theta}}d\theta$.

However, you should know that $\displaystyle \frac{1}{z}$ is analytic everywhere except where z= 0 which, with $\displaystyle |z- z_0|> 2$, means everywhere inside this contour. - Mar 1st 2010, 02:19 PMshawsend
Hi. Here's something fun to check your knowledge about contour integrals: pin the center of the unit circle at the point 2i. Now, drop it straight through the singular point of $\displaystyle 1/z$ at the origin until the center rests at -2i. Now, how does the value of the integral $\displaystyle \mathop\oint\limits_{C} \frac{1}{z}dz$ change as the circle falls?