4. MAPPING BY ELEMENTARY FUNCTIONS

    

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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http://www.malilla.supereva.it/Pages/Fractals/Cmapper/cm.html

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 = (x² – y²) + 2xyi.  That is,   (x, y) (x² – y², 2xy).

So z-points on the hyperbola x² – y² = 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'.

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Quiz 4.1

    

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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.

Notes
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

Thus

   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  (u – ¹/_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.

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QUIZ 4.2

   

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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:

.

Notes
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 z_i, w_i  (1 i 3).

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.

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QUIZ 4.3

     

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The Transformation w = exp z

As before, we set z = x + iy,  w = cis and w = exp z  gives   = e^x 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 = e^c. 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

e^a e^b,  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.

QUIZ 4.4

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