Möbius transformation and its applications
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. First, suppose c = 0. Hence S(z) = (a/d)z + (b/d) so if if S 1(z) = (a/d)z, S 2(z) = z + (b/d), then S 2 S 1 = S and we are done. Now let c 0 Then S 1(z) = z + d/c, S 2(z) = 1/z, S 3(z) = S 4 S 3 S 2 S 1(z) = S(z) = az + b cz + d. bc ad z, S 4(z) = z + a/c. c 2
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 point 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! Proof. 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. Consider 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, the cross ratio of z 1, z 2, z 3, z 4 is defined by (z 1, z 2, z 3, z 4) = (z1 z3)(z2 z4) (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 is a Möbius transformation defined by S(z) = (z, z 2, z 3, z 4) = such that S(z 2) = 1, S(z 3) = 0 and S(z 4) =. (z z3)(z2 z4) (z 2 z 3)(z z 4)
Result:The cross ratio is invariant under 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) = S(z 3) = 0 and ST 1 (Tz 4) = S(z 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)). ST 1 (T (z 1)) = S(z 1) = (z 1, z 2, z 3, z 4) = (T (z 1), T (z 2), T (z 3), T (z 4)).
Result: If z 2, z 3, z 4 are three distinct points in C and if w 2, w 3, w 4 are also three distinct points of C, then there is one and only one Möbius transformations S such that Sz 2 = w 2, Sz 3 = w 3, Sz 4 = w 4. Proof 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 has desire property. If R is another Möbius transformations. with Rz j = w j for j = 2, 3, 4 then R 1 S has three fixed poins (z 2, z 3 and z 4). Hence R 1 S = I or R = S.
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 (z, z 1, z 2, z 2) = (T (z), T (z 1), T (z 2), T (z 2)). i.e. which on solving gives (z, 1, 0, 1) = (T (z), i,, 1) T (z) = (i + 1)z + (i 1). 2z
Theorem. A Möbius transformation takes circles onto circles. 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 onto circles. To prove this result 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 u = x x 2 + y and v = y 2 x 2 + y. 2 Similarly, x = u u 2 + v and y = v 2 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 y = v ) we have u 2 +v 2 ( ( ) 2 ( u a + v ) ) 2 +b u 2 + v 2 u 2 + v 2 (i.e. substituting x = u u 2 +v 2 ( ) ( u +c u 2 + v 2 and v 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.