Complex Space : Why unit disc? 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 ? 
 A: 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.
