, Möbius Transformations Slides-13
Let γ : [a, b] C be a smooth curve in a domain D. Let f be a function defined at all points z on γ. Let C denotes the image of γ under the transformation w = f (z). The parametric equation of C is given by C(t) = w(t) = f (γ(t)), t [a, b]. Suppose that γ passes through z 0 = γ(t 0), (a < t 0 < b) at which f is analytic and f (z 0) 0. Then w (t 0) = f (γ(t 0))γ (t 0) = f (z 0)γ (t 0). That means arg w (t 0) = arg f (z 0) + arg γ (t 0).
Let C 1, C 2 : [a, b] C be two smooth curves in a domain D passing through z 0, with K 1 and K 2 their images under the transformation w = f (z). Then, arg K 1(t 0) = arg f (z 0) + arg C 1(t 0) arg K 2(t 0) = arg f (z 0) + arg C 2(t 0) that means arg K 2(t 0) arg K 1(t 0) = arg C 2(t 0) arg C 1(t 0).
Definition: A transformation w = f (z) is said to be conformal if it preserves angel between oriented curves in magnitude as well as in orientation. Note: From the above observation if f is analytic in a domain D and z 0 D with f (z 0 ) 0 then f is conformal at z 0. Let f (z) = z. Then f is not a conformal map. It preserves only the magnitude of angles between smooth curves but not orientation. Such transformations are called isogonal mapping. Let f (z) = e z. Then f is a conformal at every point in C as f (z) = f (z) = e z 0 for each z C. Let f (z) = sin z. Then f is a conformal map on C \ {(2n + 1) π 2 : n Z}.
Scale Factors Definition: If f is analytic at z 0 and f (z 0 ) = 0 then the point z 0 is called a critical point of f. Theorem: If f is analytic at z 0, f (z 0 ) = = f (k 1) (z 0 ) = 0 and f k (z 0 ) 0, then the mapping w = f (z) magnifies the angle at the vertex z 0 by a factor k. Proof. Since f is analytic at z 0 we have f (z) = f (z 0) + (z z 0)f (z 0) + 1 2! (z z0)2 f (z 0) + [ ] = f (z 0) + (z z 0) k 1 k! f k 1 (z 0) + (k + 1)! (z z0)f k+1 (z 0) + That means and = f (z 0) + (z z 0) k g(z) (say) f (z) f (z 0) = (z z 0) k g(z) arg (w w 0) = arg (f (z) f (z 0)) = k arg (z z 0) + arg g(z).
Scale Factors Let C 1, C 2 be two smooth curves passing through z 0. Let the image curves be K 1 = f (C 1) and K 2 = f (C 2). If z 1 is a variable points approaching to z 0 along C 1, then w 1 = f (z 1) will approach to w 0 = f (z 0) along the image curves K 1. Similarly if z 2 is a variable points approaching to z 0 along C 2, then w 2 = f (z 2) will approach to w 0 = f (z 0) along the image curves K 2 Let θ and Φ be the angle between C 1, C 2 at z 0 and between K 1, K 2 at f (z 0) respectively.
Scale Factors Then Φ = lim z 1,z 2 z 0 arg = lim arg z 1,z 2 z 0 = lim z 1,z 2 z 0 = kθ. ( ) f (z1) f (z 0) f (z 2) f (z 0) ( ) (z1 z 0) k g(z 1) (z 2 z 0) k g(z 2) [ z1 z0 k arg + arg g(z1) z 2 z 0 g(z 2) ]
Scale Factors Question: Find the image of the unit square S = {x + iy : 0 x 1, 0 y 1} under the map w = f (z) = z 2? Answer: The map f (z) is conformal at all z 0. So the vertices z 1 = 1, z 2 = 1 + i and z 3 = i are mapped onto right angles at he vertices w 1 = 1, w 2 = 2i and w 3 = i. But f (0) = 2 0. so the angle at the vertex z 0 is magnified by the factor 2.
For constants a, b, c, d C with ad bc 0 the function w = S(z) = az + b cz + d is called a Möbius transformation. It is also known as a bilinear transformation or a linear fractional transformation. Observation: The map S is analytic on C \ { d c }. Every Möbius transformation is a conformal map. S (z) = (cz + d)a c(az + b) (cz + d) 2 = ad bc (cz + d) 2 0. Composition of two Möbius transformations is a Möbius transformation.
S(z) = az + b cz + d The map S : C \ { d c } C \ { a c } is one one and onto. Define T : C \ { a c } C \ { d dw + b c } by T (w) = cw a. Then S T (w) = w and T S(z) = z. So, S is invertible and S 1 = T. If we define S( d c ) = and S( ) = a c, then is a bijection. S : C { } C { }
Some Möbius transformations S(z) = az + b cz + d Let a C. Then S(z) = z + a is called translation map. Let 0 a C. Then S(z) = az is called dilation map. (If a < 1, S is called contraction map and if a > 1, S is called expansion map.) Let θ R. Then S(z) = e iθ z is called rotation map. The map S(z) = 1 z is called inversion map.
Every Möbius transformation is the composition of translations, dilations and the inversion. Proof. Let w = S(z) = az + b, ad bc 0 be a Möbius cz + d transformation. If c = 0, then S(z) = (a/d)z + (b/d) and S 2 S 1 = S where If c 0, put S 1 (z) = (a/d)z, S 2 (z) = z + (b/d). S 1 (z) = z+d/c, S 2 (z) = 1/z, S 3 (z) = and get S 4 S 3 S 2 S 1 (z) = az + b cz + d = S(z). bc ad c 2 z, S 4 (z) = z+a/c,
A point z 0 C is called a fixed point of a function f if f (z 0 ) = z 0. Question: What are the fixed points of a Möbius transformation S? That is what are the point z satisfying S(z) = z. Answer: If z satisfies the condition then S(z) = az + b cz + d = z, cz 2 (a d)z b = 0. A Möbius transformation can have at most two fixed points unless it is an identity map. If c = 0, then z = b a d ( if a = d) is the only fixed point of S.
Question: How many Möbius transformation are possible by its action on three distinct points in C? Answer: One. Let S and T be two Möbius transformations such that S(a) = T (a) = α, S(b) = T (b) = β and S(c) = T (c) = γ, where a, b, c are three distinct points in C. Then, T 1 S(a) = a, T 1 S(b) = b and T 1 S(c) = c. So we have a Möbius transformation T 1 S having three fixed points. Hence T 1 S = I. That is S = T. Question: How to find a Möbius transformation if its action on three distinct points in C is given?
Definition: Given four distinct points on the z 1, z 2, z 3, z 4 C, their cross ratio is defined by (z 1, z 2, z 3, z 4 ) = (z 1 z 3 )(z 2 z 4 ) (z 2 z 3 )(z 1 z 4 ). If z 2 = then (z 1, z 2, z 3, z 4 ) = (z 1 z 3 ) (z 1 z 4 ) If z 3 = then (z 1, z 2, z 3, z 4 ) = (z 2 z 4 ) (z 1 z 4 ) If z 4 = then (z 1, z 2, z 3, z 4 ) = (z 1 z 3 ) (z 2 z 3 ) The cross ratio gives rise to a Möbius transformation defined by S(z) = (z, z 2, z 3, z 4 ) = (z z 3)(z 2 z 4 ) (z 2 z 3 )(z z 4 ). Note that S(z 2 ) = 1, S(z 3 ) = 0 and S(z 4 ) =.
Result: A cross ratio is invariant under any Möbius transformation. i.e., if z 1, z 2, z 3, z 4 are four distinct points in C and T is any Möbius transformation then (z 1, z 2, z 3, z 4 ) = (T (z 1 ), T (z 2 ), T (z 3 ), T (z 4 )). Proof. We know that S(z) = (z, z 2, z 3, z 4 ) is a Möbius transformation such that S(z 2 ) = 1, S(z 3 ) = 0 and S(z 4 ) =. Then ST 1 is a Möbius transformation such that ST 1 (Tz 2 ) =S(z 2 ) = 1, ST 1 (Tz 3 ) = 0 and ST 1 (Tz 4 ) =. Since a Möbius transformation is uniquely determined by its action on three distinct points in C we have So ST 1 (z) = (z, T (z 2 ), T (z 3 ), T (z 4 )). (T (z 1 ), T (z 2 ), T (z 3 ), T (z 4 )) = ST 1 (T (z 1 )) = S(z 1 ) = (z 1, z 2, z 3, z 4 ).
Problem: For given two triplets {z 2, z 3, z 4 } and {w 2, w 3, w 4 } in C, to find the Möbius transformations S such that Sz 2 = w 2, Sz 3 = w 3, Sz 4 = w 4. Solution. Let T (z) = (z, z 2, z 3, z 4 ), M(z) = (z, w 2, w 3, w 4 ) and put S = M 1 T. Clearly S is the desired transformation. Note that we must have (z, z 2, z 3, z 4 ) = (S(z), w 2, w 3, w 4 ). One gets S solving this equation.
Question: Find a Möbius transformation that maps z 1 = 1, z 2 = 0, z 3 = 1 onto the points w 1 = i, w 2 =, w 3 = 1. Answer: We know that Thus, which on solving gives (z, z 1, z 2, z 2 ) = (S(z), S(z 1 ), S(z 2 ), S(z 2 )). (z, 1, 0, 1) = (S(z), i,, 1) S(z) = (i + 1)z + (i 1). 2z
Theorem. A Möbius transformation takes a circle or a line onto a circle or a line. Proof. Recall that every Möbius transformation is the composition of translations, dilations and the inversion. It is easy to show that translations, dilations takes circles/lines onto circles/lines. To complete the proof it is enough to show that the transformation inversion takes circles onto circles. Consider the mapping w = S(z) = 1 z = z z 2. If w = u + iv and z = x + iy then x u = x 2 + y 2 and v = y x 2 + y 2. Similarly, x = u u 2 + v 2 and y = v u 2 + v 2.
The general equation of a circle is a(x 2 + y 2 ) + bx + cy + d = 0. (1) Applying transformation w = 1 z (i.e. substituting x = u and y = v ) we have u 2 +v 2 ( ( ) 2 ( u a + v ) ) 2 ( ) ( u +b +c v u 2 + v 2 u 2 + v 2 u 2 + v 2 u 2 + v 2 u 2 +v 2 ) +d = 0 Which on simplification becomes d(u 2 + v 2 ) + bu cv + a = o (2)
Find a Möbius transformation that takes UHP to RHP. Find the image of unit disc under the map w = f (z) = Ans: Re w > 1 2. Find the image of D = {z : z + 1 < 1} under the map w = f (z) = (1 i)z + 2 (1 + i)z + 2. z 1 z.