There are many practical situations where we try to simplify a problem by transforming it. We investigate how various regions in the plane are transformed by elementary analytic functions.

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Linear Functions (I)

1. w = f(z) = z + c  (c = (c1, c2) C) – this is a translation through c.

Each z = (x, y) maps to w = (x + c1, y + c2).

2. Let b = | b | cis be a complex constant and let w = f(z) = bz.

If z = r cis , then w = r.| b | cis( + ).
That is, each point (r,
) maps to (| b | r, + ).

Thus we have a rotation about O through and an enlargement through | b |.

Linear Functions (II)

The general linear function is the mapping w = f(z) = bz + c.
This is the rotation and enlargement (2), followed by the translation (1)
previously discussed. That is,

z bz bz + c.

Example  Consider the transformation w = iz + (i + 1).

This maps square A to square B in the adjacent figure.
Notice that S is an invariant point of this transformation.
Geometrically, we have:                     

rotation translation = rotation.

The Function f(z) = z2 (I)

Let w = f(z) = z 2. Then if z = r cis and w = cis , we have

cis = r2 cis 2  and  (r,) (, ) = (r2,2).

Example The first quadrant of the complex plane maps to the top half-plane under this mapping, since the angle range 0 /2 maps to the range 0 .

We note that in general, the mapping w = f(z) = z2 will not be 1–1. For example, the points z both map to the same w.

The Function f(z) = z2 (II)

Let us now write w = f(z) = z 2 in cartesian coordinates. Setting w = u + iv, z = x + iy, we obtain u + iv = (x2y2) + 2xyi.  That is,   (x, y) (x2y2, 2xy).

So z-points on the hyperbola x2y2 = k map to points on the straight line u = k.
Similarly z-points on the hyperbola 2xy = k' map to points on the straight line v = k'.



The Function f(z) = 1/z

The mapping w = f(z) =1/z (equivalently z = 1/w) sets up a 1–1 correspondence between points in the z- and w-planes excluding z = 0, w = 0.

In polar coordinates w = 1/z becomes cis = 1/r cis (–).

This transformation is the product of two simpler transformations:

z = r cis z' = 1/r cis  and z' = 1/r cis w = '  

inversion in the unit circle followed by reflection in the x-axis.

1. Under inversion, the points on
remain invariant.

2. Under w = 1/z, (and under inversion),

  • the centre of remains invariant;
  • lines through O map to lines through O;
  • the interior of the exterior of ;
    (in fact, circles centred at O map to circles centred at O).

Mapping Circles and Lines under f(z) = 1/z (I)

Question What is the effect of f(z) = 1/z on more general lines and circles?

Expressing w = f(z) = 1/z in terms of cartesian coordinates, we obtain


   and inversely    

Mapping Circles and Lines under f(z) = 1/z (II)

Now consider the equation

a(x2 + y2) + bx + cy + d = 0 (a, b, c ,d R). (*)

If a 0, this is the equation of a circle; if a = 0, the equation of a straight line.

Substituting for x, y in terms of u, v we get

d(u2 + v2) + bu – cv + a = 0 (†)

Points (u, v) satisfying (†) correspond to points (x, y) satisfying (*). Hence we have the four cases:

(a) a 0, d 0. Circle not through O circle not through O.

(b) a 0, d = 0. Circle through O straight line not through O.

(c) a = 0, d 0. Straight line not through O circle through O.

(d) a = 0, d = 0. Straight line through O straight line through O.

Example of a Mapping under f(z) = 1/z

What is the image under w = f(z) = 1/z of line x = c?

The line x = c maps to the set of points satisfying

    that is,    u2 + v2 u/c = 0,   or  (u1/2c)2 + v2 = (1/2c)2.

This is the circle with centre (1/2c, 0), passing through the origin O.

The half plane x > c maps to the interior of the disc.

QUIZ 4.2  


Bilinear Transformations

Let T be the transformation , (ad – bc 0, a, b, c, d C).

This is called a bilinear transformation as it can be rewritten as

cwz + dw – az – b = 0

– an equation which is linear in each of its variables, z and w.

Solving for w in terms of z, we see that the inverse of T is another bilinear transformation:


1.  The singular points for T, T–1 are z = –d/c, w = a/c respectively.
As each mapping has just one singular point, we write

z = –d/c w = ’, w = a/c ‘z = ’.

2. The set of bilinear transformations forms a group.

Understanding the Bilinear Transformation

In the formula for a bilinear transformation, if c 0, we can write

Setting z' = cz + d,  z'' = 1/z', we obtain


If c = 0, we get an expression of type (*) immediately.

It follows that any bilinear transformation can be obtained as the composition of linear transformations and the mapping f(z) = 1/z , all of which map lines and circles to lines and circles.

Hence any bilinear transformation maps the set of lines and circles to itself.

Images under a Bilinear Transformation

The following reult will be useful.

Theorem 4.1 There exists a unique bilinear transformation which maps three given distinct points z1, z2, z3 onto three distinct points w1, w2, w3 respectively.

Proof The algebraic expression for the bilinear transformation can be written as
cwz + dw – az – b = 0. This is an equation in four unknowns a, b, c, d. Hence the three ratios a : b : c : d of these numbers are determined by substituting three pairs of corresponding values of zi, wi  (1 i

In practice, this means that in general if we allocate image points to three points in the z-plane, then the bilinear transformation will be completely determined. Conversely, if we want to find the image of a circle (say) under a given bilinear transformation, then it is sufficient to find the images of three points on the circle.

Bilinear Transformation: Example

Let a C be constant with Im a > 0. Find the image of the upper half plane (y 0) under the bilinear transformation


We first consider the boundary. For z on x-axis, we have | z – a | = | z – |.
Hence for such points, .

That is, the x-axis maps to the unit circle having centre O.

Also, z = a (a point in the upper half plane) maps to w = 0 (a point interior to the unit circle). Observing that the mapping is continuous, we deduce that the image of the upper half plane is the interior of the disc.


The Transformation w = exp z

As before, we set z = x + iy,  w = cis and w = exp z  gives   = ex and = y.

The line y = c maps onto the ray = c (excluding the end-point O) in a 1–1 fashion. Similarly, the line x = c maps onto the circle = ec. However, here, an infinite number of points on the line map to each image point.

Combining these results, we see that the rectangular region  

a x b,  c y d

is mapped to the region

ea eb,  c d

bounded by portions of circles and rays.

This mapping is 1–1 if  d – c < 2.

A Special Case of w = exp z

As a particular case, the strip 0 < y < maps to the upper half plane > 0, 0 < < of the w-plane.

It is interesting to map the boundary points here. This mapping is useful in applications.

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