Mathematics Spring 2007

INDIAN INSTITUTE OF TECHNOLOGY BOMBAY Department of Mathematics Spring 2007

MA 204 - Comp. Anal. & PDE’s Tutorial No.9

Q.1-L Prove that every linear, fractional transformation (L.F.T.) is a composite of a translation, a rotation, a dilation (or a contraction) and an inversion. az + b solution: Let T (z) = where ad − bc =0.6 If c =6 0, T (z) can cz + d bc − ad 1 a be written as 2 d + which is a composite of a translation c z + c c (by d/c), the inversion (w.r.t. the origin), a rotation and a dilation (or a contraction) and ﬁnally the translation by a/c. If c = 0, then a b d =6 0 and T (z) becomes z + which is a composite of a rotation, d d a dilation (or a contraction) and a translation.

Q.2-L Let z1, z2, z3, z4 be four distinct, extended complex numbers. z1 − z3 z2 − z4 Deﬁne their cross ratio to be (z1, z2, z3, z4) = . z1 − z4 z2 − z3 Modify this deﬁnition suitably if any of the points is ∞. Note that the order of the points matters. Prove that the cross ratio is preserved under linear fractional transformations.

Solution: Whenever any one of z1, z2, z3, z4 is ∞, the ratio of the two terms in which it occurs is to be treated as 1. This is justiﬁed because that is exactly the limit of the cross ratio as the corresponding zk tends to ∞. For example, z1 − z3 z2 − z4 (z1, z2, ∞, z4) =z lim 3→∞ z1 − z4 z2 − z3 z2 − z4 z1 − z3 = zlim z1 − z4 3→∞ z2 − z3 z2 − z4 = (1) z1 − z4

1 azk + b Now let wk = . Then (w1,w2,w3,w4) can be calculated with czk + d direct substitution and comes out as (z1, z2, z3, z4) as can be verified by brute force. A better approach is to use the last problem. Then it suffices to show that the cross ratio is preserved under each of the ’basic’ transformations, viz. a translation, a rotation, a dilation (or a contraction) and finally the inversion w.r.t. the origin. All these verifications are trivial except possibly in the case of the inversion w =1/z. In this case we have

w1 − w3 w2 − w4 (w1,w2,w3,w4) = w1 − w4 w2 − w3 1 1 1 1 z1 − z3 z2 − z4 = 1 1 1 1 z1 − z4 z2 − z3 z3 − z1 z4 − z2 = z4 − z1 z3 − z2 z1 − z3 z2 − z4 = z1 − z4 z2 − z3 = (z1, z2, z3, z4) (2)

In this derivation it is tacitly assumed that none of the zk’s equals 0 or ∞. These degenerate cases are handled by taking limits of the non-degenerate ones.

Q.3-S For every complex number w other than 1, 0, and ∞, prove that w = (w, 1, 0, ∞). Hence show that given any three distinct complex numbers z2, z3, z4, the unique L.F.T. which takes them to 1, 0 and ∞ respectively is given by T (z) = (z, z2, z3, z4). View Theorem 2 on p. 694 in Kreyszig in this light. Solution: The ﬁrst part follows from the very deﬁnition of (w, 1, 0, ∞) w − 0 as . Now, suppose z2, z3, z4 are three distinct complex numbers. 1 − 0 Let T be the unique L.F.T. which maps them to 1, 0 and ∞ respec-

2 tively. Then by preservation of cross ratios,

w = T (z) = (T (z), 1, 0, ∞)

= (z, z2, z3, z4) z − z3 z2 − z4 = (1) z − z4 z2 − z3 Theorem 2 on p. 694 is merely another way of stating that the cross ratios are preserved under L.F.T.’s.

Q.4-T Prove that every L.F.T. maps a circle onto a circle, where a ’circle’ means either a straight line or a circle. [Hint: Use Problem 1 above with Problem 10 in Tutorial 8.] Solution: In view of Problem 1, it suﬃces to prove this for each of the basic L.F.T.’s. It is trivial that all translations, rotations and dilations map straight lines into straight lines and (ordinary) circles into circles. The case of the inversion T (z)=1/z was covered by Problem 10 of Tutorial 8. Speciﬁcally, under this mapping, the image of a ’circle’ is a straight line or a circle depending upon whether the ’circle’ passes through the origin or not.

Q.5-S Prove that four distinct complex numbers z1, z2, z3, z4 lie on a ’circle’ if and only if the cross ratio (z1, z2, z3, z4) is a real number.

Solution: Let C be the unique ’circle’ passing through z2, z3 and z4. Let T be the (unique) L.F.T. which takes z2, z3, z4 to 1, 0, ∞ respectively and let w1 = T (z1). Then by Problems 2 and 3,

(z1, z2, z3, z4) = (w1, 1, 0, ∞) = w1 (2)

−1 Since T as well as T takes ’circles’ to ’circles’, it follows that z1 ∈ C if and only if w1 lies on the real line (which is the unique ’circle’ passing through 1, 0 and ∞). So, by (1), z1, z2, z3, z4 are ’concyclic’ if and only if w1, i.e. (z1, z2, z3, z4) is real.

3 ∗ Q.6-T Let z2, z3, z4 be distinct points on a ’circle’ C in IC . Two ∗ ∗ points z and z are said to be symmetric w.r.t. C if (z , z2, z3, z4) = ∗ (z, z2, z3, z4). Prove that if C is a straight line then z and z are symmetric w.r.t. C if and only if they are reflections of each other into C while if C is an (ordinary) circle with centre M and radius r and P,Q are points represented by z, z∗ respectively then z, z∗ are symmetric w.r.t. C if and only if P,Q lie on the same ray from M and MP.MQ = r2. (As a consequence, it follows that symmetry is independent of the choice of the three points z2, z3, z4 on the ’circle’.) Solution: Assume first that C is a straight line. Then by a translation and a rotation, we can convert C to the real axis IR. Re- flections are preserved under both these transformations. Moreover these transformations are also L.F.T.’s. So they preserve cross ratios. Now, for the real line, points representing z and z∗ are reflections of each other if and only if z∗ = z. Thus the problem is reduced to showing that if z2, z3, z4 are any real numbers then for any complex ∗ ∗ ∗ numbers z and z , z = z if and only if (z , z2, z3, z4) = (z, z2, z3, z4). This is best done by a direct calculation. z − z3 z2 − z4 (z, z2, z3, z4) = (1) "z − z4 z2 − z3 # z − z3 z2 − z4 = (2) z − z4 z2 − z3 z − z3 z2 − z4 = (3) z − z4 z2 − z3 = (z, z2, z3, z4) (4) where the transition from (1) to (2) is by properties of complex con- jugation while that from (2) to (3) follows from the assumption that z2, z3, z4 are real. ∗ The problem now reduces to showing that (z , z2, z3, z4)=(z, z2, z3, z4) if and only if z∗ = z. But this follows from the fact that for any

4 complex number w, the cross ratio (w, z2, z3, z4) is the image of the unique L.F.T. which takes z2, z3, z4 to 1, 0, ∞ respectively. (Note that if luckily z2, z3, z4 were 1, 0, ∞ respectively, then we would get an easy proof by Problem 3. But we have no way to ensure this. All we know is that they are real and hence equal their respective complex conjugates. So, we have to go through (4).) The argument in case C is an ordinary circle is similar in principle but diﬀers in details. By a translation and a dilation we may suppose that C is the unit circle. Since reﬂections as well as cross ratios are invariant under these transformations, the problem is reduced to showing that if z2, z3, z4 are any three distinct points on ∗ the unit circle, then (z , z2, z3, z4) = (z, z2, z3, z4) holds if and only if ∗ z =1/z. Once again the proof begins by recasting (z, z2, z3, z4). The steps are strikingly similar to those in the derivation of (4) above, except that instead of zk = zk for k =2, 3, 4 we now have zk =1/zk. With this change, instead of (3) we now get

z3z − 1 z4 − z2 (z, z2, z3, z4) = (5) z4z − 1 z3 − z2

By a direct computation, the R.H.S. equals (1/z, z2, z3, z4). The rest of the argument is the same as that when C is a straight line.

Q.7-S Prove that symmetry is preserved under L.F.T.’s. Solution: This is immediate because symmetry is deﬁned in terms of cross ratios and the cross ratios are preserved under L.F.T/’s. (Actually, the solution to the last problem already used a special case of this, viz. invariance of symmetry under translations, rotations and dilations.)

Q.8-L Prove that every L.F.T. which maps the open unit disc onto z − a itself is of the form T (z) = c for some complex numbers 1 − az a, c with |a| < 1 and |c| =1.

5 Solution: Let T be an L.F.T. which takes the open unit disc D onto itself. Let C be the unit circle which is the boundary of D. Then we must have T (C) = C. (This is a topological fact which is intuitively obvious. A proof uses the concept of closure of a set and the fact that both T and T −1 are continuous. We omit such proofs.) Now, let a = T −1(0). Then a ∈ D. We assume a =6 0 first. The reflections of 1 a and 0 in C are and ∞ respectively. So, because of preservation a 1 of symmetry under L.F.T.’s, T must map to ∞. So(z − a) is a 1 a factor of the numerator and (z − a) is a factor of the denominator of T (z). Hence T (z) is of the form z − a T (z) = A 1 (1) z − a for some constant A. Multiplying the denominator by −a we can recast this as z − a T (z) = c (2) 1 − az for some constant c (possibly different from the earlier constant A). We still have to show that |c| = 1. For this we use that |T (z)| = 1 |1 − a| whenever |z| = 1. In particular, taking z = 1, we have |c| =1 |1 − a| which shows that |c| = 1, since 1−a and 1−a are complex conjugates of each other and hence have the same magnitude. The case a = 0 still remains. In this case we have T (0) = 0 whence the numerator of T (z) is a constant multiple of z. Also, ∞ is the reflection of 0 in C. Hence T (∞) = ∞ which means the denominator of T (z) is a constant. Put together, T (z) = cz for some complex number z. Since T (z)| = 1 whenever |z| = 1, we have |c| = 1. Conversely, it is easy to see that every L.F.T. of the form (2) with |c| = 1 maps C to itself. For, if |z| = 1 then |z − a| |T (z)| = |c| |1 − az|

6 |z − a| = |z|(|1 − az)| |z − a| = |z − a| =1 (3)

Further T (a)=0. So |a| < 1 ensures that it is the inside and not the outside of C that is mapped to the inside of C.

1 1 Q.9-T Map the region between the circles |z| = 1 and |z − 2| = 2 conformally onto an inﬁnite strip and then onto a half plane. Solution: (Given separately.)

Q.10-S Prove that an entire function whose real part is bounded below is a constant function. How will you generalise this result? Solution: Suppose f(z) = u+iv is an entire function with u(z) ≥ A (say) for all z ∈ ICwhere A is a ﬁxed real number. Let D be the closed half-plane {(u,v): u ≥ A}. Then f maps ICinto D. Let T bea linear fractional transformation which maps D onto the (closed) unit disc. Then T is holomorphic on D and hence the composite function g(z) = T (f(z)) is an entire function. But, by very construction, it is bounded. So, by Liouville’s theorem, g(z) is a constant. But then so is T −1 ◦ g, which is precisely f. More generally, an entire function cannot be bounded in any direction. That is, given any complex number ξ with |ξ| = 1, and any λ > 0, the function must assume as a value a complex number of the form µξ for some µ > 0. For, if this is not so, then f(z) will take values in the closed half plane bounded by a line through λξ and perpendicular to ξ. Composing f with an L.F.T. which trans- forms such a half plane to the unit disc will give an entire, bounded function.

7 Still more generally, the argument will apply to show that if D is any domain which can be mapped conformally and bijectively onto the open unit disc then an entire function taking values only in D is a constant. The question, naturally, is to characterise all such domains. There is a famous theorem, called the Riemann Mapping Theorem which asserts that every simply connected domain other than the complex plane ICitself, is conformally equivalent to the open unit disc. But the proof is far from trivial.

Q.11-S Map the region bounded in the ﬁrst quadrant bounded by the coordinate axes and the hyperbola y =1/x conformally onto the upper half plane. Solution: (Given separately.)

Q.12-O Besides the Riemann sphere, there is another interpretation of the extended complex plane IC∗ called the complex projec- tive line. It makes it easier to see what is really ’linear’ in an L.F.T. Consider an ordered pair (z1, z2) of (ordinary) complex numbers z1, z2 at least one of which is non-zero. If z2 =06 , we associate the complex number z1/z2 to this pair. Other- wise we associate ∞. Note that the same complex number may be associated to many diﬀerent pairs. Now suppose an extended complex number z corresponds to the pair (z1, z2). Then for any complex numbers a,b,c,d with ad =6 bc, we have az + b az1 + bz2 w1 w1 a b z1 = = where   =    . This is a cz + d cz1 + dz2 w2 w2 c d z2       well-deﬁned, non-singular linear transformation    of the complex two dimensional vector space into itself.