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. You may find the following website helpful: http://www.malilla.supereva.it/Pages/Fractals/Cmapper/cm.html 1. w = f(z) = z + c (c = (c_{1}, c_{2}) C) – this is a translation through c. Each z = (x, y) maps to w = (x + c_{1}, y + c_{2}). 2. Let b =  b  cis be a complex constant and let w = f(z) = bz. If z = r cis , then w = r. b  cis( + ). Thus we have a rotation about O through and an enlargement through  b . The general linear function is the mapping w = f(z) = bz + c. z bz bz + c.
Let w = f(z) = z ^{2}. Then if z = r cis and w = cis , we have cis = r^{2} cis 2 and (r,) (, ) = (r^{2},2).
We note that in general, the mapping w = f(z) = z^{2} will not be 1–1. For example, the points z both map to the same w. Let us now write w = f(z) = z ^{2} in cartesian coordinates. Setting w = u + iv, z = x + iy, we obtain u + iv = (x^{2} – y^{2}) + 2xyi. That is, (x, y) (x^{2} – y^{2}, 2xy). So zpoints on the hyperbola x^{2} – y^{2 }= k map to points on the straight line u = k. The mapping w = f(z) =^{1}/_{z} (equivalently z = ^{1}/_{w}) sets up a 1–1 correspondence between points in the z and wplanes 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 xaxis. Notes 2. Under w = ^{1}/_{z}, (and under inversion),
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 Now consider the equation a(x^{2 }+ y^{2}) + 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(u^{2 }+ v^{2}) + 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. 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, u^{2 }+ v^{2 }– ^{u}/_{c }= 0, or (u – ^{1}/_{2}_{c})^{2 }+ v^{2 }= (^{1}/_{2}_{c})^{2}. This is the circle with centre (^{1}/_{2}_{c}, 0), passing through the origin O. The half plane x > c maps to the interior of the disc. 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 z = –^{d}/_{c} ‘w = ’, w = ^{a}/_{c} ‘z = ’. 2. The set of bilinear transformations forms a group. 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. The following reult will be useful. Theorem 4.1 There exists a unique bilinear transformation which maps three given distinct points z_{1}, z_{2}, z_{3} onto three distinct points w_{1}, w_{2}, w_{3} respectively. Proof The algebraic expression for the bilinear transformation can be written as In practice, this means that in general if we allocate image points to three points in the zplane, 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. 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 xaxis, we have  z – a  =  z – . That is, the xaxis 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. 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 endpoint 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. As a particular case, the strip 0 < y < maps to the upper half plane > 0, It is interesting to map the boundary points here. This mapping is useful in applications.
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