Conformal Mappings - Complex Analysis, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

Conformal Mapping

We shall show now that the curves

u(x, y)= constant and v(x, y)= constant

intersect each other at right angles (we say that they are orthogonal). To see this we note that along the curve u(x, y)= constant we have du =0. Hence

Thus, on these curves the gradient at a general point is given by

Similarly along the curve v(x, y)= constant, we have

The product of these gradients is

where we have made use of the Cauchy-Riemann equations. We deduce that the curves are orthogonal.

As an example of the practical application of this work consider two-dimensional electrostatics.
If u = constant gives the equipotential curves then the curves v = constant are the electric lines of force. Figure 1 shows some curves from each set in the case of oppositely-charged particles near to each other; the dashed curves are the lines of force and the solid curves are the equipotentials.

In ideal fluid flow the curves v = constant are the streamlines of the flow.

In these situations the function w = u +iv is the complex potential of the field.

Conformal Mapping

A function w = f (z) can be regarded as a mapping, which ‘maps’ a point in the z-plane to a point in the w-plane. Curves in the z-plane will be mapped into curves in the w-plane.
Consider aerodynamics. The idea is that we are interested in the fluid flow, in a complicated geometry (say flow past an aerofoil). We first find the flow in a simple geometry that can be mapped to the aerofoil shape (the complex plane with a circular hole works here). Most of the calculations necessary to find physical characteristics such as lift and drag on the aerofoil can be performed in the simple geometry - the resulting integrals being much easier to evaluate than in the complicated geometry.

Consider the mapping

w = z².

The point z = 2+i maps to w =(2 + i)² = 3+ 4i. The point z = 2+i lies on the intersection of the two lines x =2 and y =1.To what curves do these map? To answer this question we note that a point on the line y =1 can be written as z = x +i. Then

w =(x +i)² = x2 − 1+2xi

As usual, let w = u +iv, then

u = x² − 1 and v =2x

Eliminating x we obtain:

4u =4x² − 4= v² = 4 or v² = 4+4u.

Note that the product of the gradients at (3,4) is −1 and therefore the angle bewteen the curves at their point of intersection is also 90⁰. Since the angle bewteen the lines and the angle between the curves is the same we say the angle is preserved.

In general, if two curves in the z -plane intersect at a point z₀, and their image curves under the mapping w = f (z)intersect at w₀ = f (z₀) and the angle between the two original curves at z₀ equals the angle between the image curves at w0 we say that the mapping is conformal at z0.

An analytic function is conformal everywhere except where f '(z)=0.

Inversion

The mapping

is called an inversion.It maps the interior of the unit circle in the z -plane to the exterior of the unit circle in the w-plane, and vice-versa. Note that

 and similarly

so that

A line through the origin in the z-plane will be mapped into a line through the origin in the w-plane. To see this consider the line y = mx, for m constant. Then

so that v = −mu, which is a line through the origin in the w-plane.

Similarly, it can be shown that a circle in the z -plane passing through the origin maps to a line in the w-plane which does not pass through the origin and a circle in the z-plane which does not pass through the origin maps to a circle in the w-plane which does not pass through the origin. The inversion mapping is an example of the bilinear transformation:

  where we demand that ad − bc ≠ 0

(If ad − bc =0 the mapping reduces to f (z)= constant).