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This may sound a newbie question anyway I'm yet new to this area,

In complex space for simplicity the properties of the functions on curves are sometimes considered on the unit circle on complex plane, with the center (0, 0) . So basically I would like to know why until circle ? Does it mean that all other smooth curves can be represented by combinations of circles with arbitrary radius ?

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Frequently the specific shape of the curve is not important, and a circle is nicely parametrized. –  copper.hat Sep 10 '12 at 22:05
    
Why it is not important ? probably for particular function there might be a special point somewhere not on unit circle. I mean for example for "simplicity" Laplas's problem is considered on unit circle and so on and the corresponding solutions are provided in case the contour is unit circle. Please provide some more info why the contour shape is not important ? –  deimus Sep 10 '12 at 22:10
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Very loosely, the path integral over two homotopic curves are equal. So a circle is a convenient example of a curve of index 1 around $0$. The Riemann mapping theorem is another. –  copper.hat Sep 10 '12 at 22:28

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Terminological remark: it's best to use different words for the unit circle $|z|=1$ and the unit disk $|z|<1$. When you mention the Laplace equation, you probably have the disk in mind.

One attractive feature of the circle $|z|=1$ is that it's a group under multiplication. As a byproduct of the group structure, $|z|=1$ has a transitive group of isometries, which is a fancy way of saying that all points look the same because we can rotate the circle. Hence, it supports a canonical rotation-invariant measure, the Lebesgue measure on the circle. Using the invariance of the Laplace operator under rotations, we can conclude (at least heuristically) that the solution of the Dirichlet problem $\Delta u=0$, $u_{\partial D}=g$ will satisfy the mean value property: $u(0)$ is the average of $g$. This may not look like much, but if we also use the invariance of the Laplacian under conformal maps (specifically the Moebius transformations of the disk), the solution of the Dirichlet problem is obtained at once.

For domains of other shape (e.g. triangle) solving the Dirichlet problem is not nearly as easy, because the above argument does not apply.

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