Conformal mapping in Complex Analysis

1. Apr 13, 2005

rhia

I would appreciate if someone could explain Conformal Mapping using Complex Analysis using an example. I get the rough idea but have no clue how complex analysis comes into the picture.

Thank You!

2. Apr 13, 2005

matt grime

3. Apr 16, 2005

rhia

Can you give an example to show how it works?
I am finding it difficult to visualize.

4. Apr 17, 2005

matt grime

No, I cannot since I cannot understand what you need to do in order to understand it. A map is conformal at a point if its derivative doesn't vanish. It is a definition. Look up how angles transform (in terms of derivatives) to see why. I've no idea what you mean ny "how it works", sorry.

The map z goes to az+b (a=/=0) is a conformal mapping (ie map conformal at all points of its domain) of C to C. it's a rotation, scaling and translation, obviously it preserves angles.

5. Apr 17, 2005

mathwonk

using complex analysis one can prove that all conformal isomorphisms of the "complex plane", extend to complex automorphisms of the extended complex plane (the compelx projective "line"), hence have form (az+b)/(cz+d).

For example, the map sending z to w = (z-i)/(z+i) is an isomorphism of the extended complex plane, which sends the points z which are closer to i than to -i, to those points w of norm less than one.

I.e. this is an isomorphism from the upper half plane, onto the open unit disc.

Examples of conformal mappings seldom use complex analysis, but proofs that they have a certain form do so.

(I am using the word conformal here in the sense of not just angle preserving, which is the correct meaning, but also orientation preserving, hence complex holomorphic.)

6. Apr 28, 2005

Starship

From the complex point of view, the porpuse is to investigate in more general terms the character of transforms for which the mapping function $$w = u(x,y) + i\nu(x,y)$$ is analytic.

Since w = f(z) is analytic, substituting into the jacobian determinant, we get

$$J \left ( \frac{u,\nu}{x,y} \right ) = \begin{vmatrix} \frac{\partial u}{\partial x} & - \frac{\partial \nu}{\partial x} \\ \frac{\partial \nu}{\partial x} & \frac{\partial u}{\partial x} \end{vmatrix} = |f'(z)|^2$$

From here there are 4 theorems which can be deduced...

Last edited: Apr 29, 2005