Module-3: Bilinear Transformation - Normal Form Fixed Points or Invariant Points The points which coincide with its transformation are called fixed or invariant points of the transformation. If z is a fixed point of the transformation 7, then T(z) The identity transformation J(2) = 2 has every 2}€ GU {oo} as its fixed points. The fixed points of the transformation w = 2? are solutions of Theorem 1. Prove that in general there are two values of 2 for which w = = but there is only one if (a —d)?+ Abe =0. Show also that if there are distinct invariant points p and q, then the transformation can be written in the form and that if there is only one invariant point p, the transformation can be written in the form 1 . 2) wp wre k is a constant. Proof. Let w , ad—be #0 (3) be a bilinear transformation. For fixed points we have ie. c2?—(a—d)z—b =0. (4) 2In general (4) gives two values of z and hence two fixed points. When (a~d)?+4be = 0, there is only one fixed point. Let p and q be two fixed points of the transformation (3). Then from (4) we obtain cp? —ap =b—dp and cq? —aq =b—dq. . az+b (a= cp)z + (b= dp) Now w—p= —p = Gr@etb-dp) oe OP eed td (a= pz =) td There fore, a =o} where k If there is only one fixed point then by (A). ce+d cz+a—2p Therefore, ————— = i Gale ~@—ae—a) e(z =p) + (a- cp) 1 EEE kp (a~ @p)(z—p) where k = This proves the theorem. a Remark 1. Equations (1) and (2) are known as the normal form or canonical form of « bilinear transformation. Remark 2. A bilinear transformation w = f(z) with more than two fied points in the extended complex plane must be the identity transformation. Note 1. (i) A bilinear transformation having only one fixed point is called a parabolic transformation and if the fired point is p then the transformation is of the formwhere k is a constant. (ii) A bilinear transformation having two fixed points p and q can be written in the form wr > (54) where k is a constant. If| k|=1, k #1, then it is called elliptic transformation and if k > 0 is real, then it és called hyperbolic transformation. (iii) A bilinear transformation which is neither parabolic nor elliptic nor hyperbolic. is called loxodromic. That is it has two fixed points and satisfies the condition k = ae, a # 0,071. Example 1. Find all the bilinear transformations which have fixed points 1 and —1. Solution. Let az+b s@) = 24 ad be £0 be the required bilinear transformation. Now fl) =1 3 atb seta and f(=1) =-1 sa-b =d-c Solving we get a = and-b =e. Therefore, az+b PES) ra @-P 40 is the required transformation. Sit Example 2. Find the normal form of the bilinear transformation w = 82 Solution. For fired points of the transformation we have _3iz41 ie. (2-1? =0 ie. 2 =i So the given transformation has only one fixed point and hence its normal form is 1 1 tk = where k is a constant.Example 3. Find the bilinear transformation with fived points —1 and 1 carrying the point 2 =i onto w Solution. Since the transformation has two distinct fired points, its normal form is where k is some constant. Since it maps = =i onto w =—i we have ~i+1 | (4) Solving we get k = 1. Hence the required transformation is wtl (zt wl \z=1 1 ie. w =<. z We now develop the specific bilinear transformation which maps three distinct points in the extended z-plane onto three distinct points in the extended w-plane. For that we first introduce the concept of cross ratio. Cross Ratio For three distinet complex numbers 1, 22, in CU {oo}, the cross ratio of four points 2, 21, 22, % is defined as (au a 29) = Goes 2) If one of the numbers is replaced by infinity, say 23, then Note 2. The cross ratio will change if the order of the factors is changed. Since four letters 21, 22, 23, 24 can be arranged in 4! = 24 ways, there will be 24 cross ratios but as a matter of fact there will be only six distinct cross ratios. The six distinct cross ratios are (21, #2, 23, 24)s (21, 2 Zu, 23), (21, 23) 225 Ha); (21, 235 2ay 22), (21, 2a, 225 23), (21, 2a, 23 22). Theorem 2. A bilinear transformation leaves a cross ratio invariant.Proof. Let az+b Tp a-re#0 (5) be abilinear transformation, Let wy, wz, wy, w1 be the images of the points 2, 2, 3, 21 respectively under the transformation (5). ‘To prove the theorem we have to show that (wi, wa, ws, we) We have (wy = w2)(ws = tea) (tor, tay tos, wa) = (we — ws)(w4 — 1) az tb +b _ (ad —be)(21 — 2 mid ~ (etal Now w;—w2 = cata Similarly, (ad = be)(z2 — 23) Ca + A(ez3 +d) (ad ~ be) (23 = 24) Oe Cag + Dcza + dy" sop cum, — (td bees ==) + d)(exi +4) Hence we obtain (= a)(en- a) _ (wr, wa wae) = E This proves the theorem. o Theorem 3. Given three distinct points, , and 25 in the extended 2-plane and three distinct points w1, wp, and wy in the extended w-plane, there exist a unique bilinear transformation w = f(z) such that f(z.) = wp for k =1, 2,3. Proof. We assume that none of the six points is oo. Let az+b cetd w =f(2) ad —be £0 be the bilinear transformation. We wish to solve for a, b, ¢, and d in terms of 21, 22) 2 wi, w2, and wy. For k =1, 2, 3 we have aztb az%+b _ (ad ~be)(2~ 2) Oo +d cytd (2+d(ex +d) (6) 6So from (6) we obtain wow, _ (extd wm ~ (S4)( © Replacing = by 22 and w by wy in (7) we obtain w2 — ws ww, ®) Multiplying (7) and (8) we have = wi)(w2 = (2 - a)(z2 — 23) 0) (2-01) ~ 25) eA)" Solving for w in terms of 2 and the six points gives the desired transformation. If one of = 00, (9) would be modified by taking the limit as zy — 00. In this the points, say case, we would have (w= wi)(w, ~ ws), Ge- 2 (w i) 2 1 Now we suppose that f(2) and g(2) are both bilinear transformations that agree at three or more points in the extended complex plane. Therefore, 2,3. wees fee) = g(ze) for k =1, Thus for k =1, 2, 3 we have (Fe g)(e) =F *((z)) =F (we) and so by Remark 2, f-!og =I. This gives = g which proves the uniqueness part of the theorem. oO Example 4. Find a bilinear transformation that maps the points 2 = i, 2, —2 onto " i, 1, —L, respectively. Solution. We know that the transformation which maps the points 21, 22, and 2s in the extended 2-plane onto the points wi, wa, and ws in the extended w-plane is Here 7Therefore we obtain the required transformation as (w—i)(1+1 _ (2—#)(2+2) (w+I-a ~ (2+2)(2-1) wi i) w+D0-1) ~ +221) ie. izw+6w = 324.2% 32+ 2i ie w +6 Example 5. Let f(z) be a bilinear transformation such that floc) = 1, fi) = i and f(-i) ~i. Find the image of the unit disc {z € C:| 2 |< 1} under f(z) Solution. Let w = %#4, ad — be 0 be the required bilinear transformation. Now f(oo) =1 a/c =1 +a =c (10) ait Ml) = Soap i > (a=dit(b+e <0. (11) Also i+b $C) 4 = TE = > (a-di-W+e) (2) Solving (10), (11) and (12) we obtain Thus w = f(z) = we see that _re®@-1 _ (rcos #1) +irsin 0 re®+1 — (reos @+1)+irsin 0 =P, “P24 2reos 8+ 1 "1? + 2reos @+1 The map of a point lying on | =| = 1 lies on the imaginary avis in the w-plane because forr=1, Rew =0. Ifr <1, then Rew <0, i.e. the image of a point lying inside the unit civele | =1 lies in the left half of the w-plane.