COMPLEX NUMBERS in GEOMETRY We Identify the Set of Complex Numbers C with the Cartesian Plane, by a + Ib ∈ C Being Thought Of

COMPLEX NUMBERS IN GEOMETRY

We identify the set of complex numbers C with the cartesian plane, by a + ib ∈ C being thought of as the point (a, b) in the plane. The rich interplay between the algebraic manipulations and the Euclidean intuition gives rise to powerful applications which we intend to explore in this note. Triangles in a plain. A number of properties related to triangles can be written conveniently in terms of its complex co-ordinates. 1 • z1, z2, z3 and w1, w2, w3 are similar (in an orientation preserving way) iﬀ the determinant:

z1 w1 1

z2 w2 1

z3 w3 1 vanishes. What happens if the similarity is orientation reversing? 2 2 • If z1, z2, z3 form an equilateral triangle, either z1 + ωz2 + ω z3 = 0 or z1 + ω z2 + ωz3 = 0. • (Several other properties were mentioned in a different handout.) Affine Transformations. Most common euclidean transformations2 have a rather simple expression in the complex plane: • Translation: A transformation that takes any line to a parallel line and preserves distances, orientation. z 7→ z + a for any a ∈ C. • Rotation (about a point p): A transformation that fixes p, preserves distances, orientation. If 0 0 q 6= p is such that it maps to q , the angle ∠q pq is called as the angle of rotation. z 7→ a(z − p) + p for any a ∈ C with aa¯ = 1. Therefore angle of rotation is independent of q. • Homothety (about a point p): A transformation that fixes p, takes a line to a parallel line. If q 6= p is such that it maps to q0, the ratio |pq0|/|pq| is called as the ratio of homothety. z 7→ a(z − p) + p with a ∈ R. Therefore ratio of homothety is independent of q. • Spiral Symmetry (about a point p): A transformation that is composition of homotheties and rotations about p. z 7→ a(z − p) + p with a ∈ C. • Reflection (about line l = (x axis)): A transformation that fixes l, distance preserving, nontrivial. z 7→ z¯. • Rigid motions: A transformation that preserve distances. z 7→ az + b or z 7→ az¯ + b for any a, b ∈ C s.t. aa¯ = 1. • Similarities: A transformation that preserves length up to a fixed scalar. z 7→ az + b or z 7→ az¯ + b for any a, b ∈ C. • Linear transformation: A transformation that fixes 0, take lines to lines. z 7→ az + bz¯ for any a, b ∈ C. • Affine transformation: A transformation that takes lines to lines. z 7→ az + bz¯ + c for any a, b, c ∈ C. Exercise 1. Prove the following:

(1) Composition of a rotation with angle θ1 with another with angle θ2 is a rotation with angle θ1 + θ2 unless θ1 + θ2 is a multiple of 2π, in which case it is a translation. (2) Composition of a homothety with ratio r1 and another with ratio r2 is a homothety with ratio r1r2 unless r1r2 = 1 in which case it is a translation. What can you say about the center of the composite homothety if r1r2 6= 1? (3) Composition of spiral symmetries is either a spiral symmetry or a translation. (4) An orientation preserving similarity is either a translation or a spiral symmetry.

1 By orientation of ∠ABC we mean the sense (clockwise or counter-clockwise) of the acute angle from AB to BC 2 By a transformation of C we mean a map C → C. 1 Exercise 2. Prove the following: (1) Given a figure A congruent to another figure B, there is a rigid motion taking A to B. If the figure A (and hence B) has at least 3 non-collinear points, the rigid motion is unique. (2) Given p, q, r not collinear and p0, q0, r0 not collinear, there is a unique bijective affine transformation taking p 7→ p0, q 7→ q0 and r 7→ r0. Geometric Inequalities. Given some complex number a, b, we have the triangle inequality |a + b| ≤ |a| + |b| with equality only if 0, a, b are collinear, 0 not between a and b. While this is already available in geometry, the advantage of working with complex numbers is that a product of complex numbers is a complex number and hence we can prove more general inequalities, those involving products of lengths. Problems.

(1) A triangle A1A2A3 and a point P0 are given in the plane. We define As = As−3 for all s ≥ 4. We construct a sequence of points P1,P2,P3, ... such that Pk+1 is the image of Pk under rotation ◦ with center Ak+1 through angle 120 clockwise (for k = 0, 1, 2, ... ). Prove that if P2013 = P0, then the triangle A1A2A3 is equilateral. (2) Consider equilateral triangles ABC and A0B0C0, both in the same plane and having the same orientation. Show that the segments AA0, BB0, CC0 can be the sides of a triangle. (3) Given three nonintersecting circles of different radii, draw the intersection of the external tangents to each pair of the circles. Show that these three points are collinear. (4) Three circles C1,C2,C3 of equal radii r have a common point O. Circles C1 and C2, C2 and C3, C3 and C1, meet again at points A, B, C respectively. Prove that the circumradius of the triangle ABC is also equal to r. (5) On each side of a parallelogram a square is drawn external to the figure. Prove that the centers of the squares are the vertices of another square. (6) Let A1A2A3 be non-equilateral and consider the points B1,B2,B3 such that triangles A1A2B3, A2A3B1 and A3A1B2 are similar with the same orientation. Then prove that triangle B1B2B3 is equilateral if and only if triangles A1A2B3, A2A3B1, and A3A1B2 are isosceles with the bases ◦ A1A2,A2A3, and A3A1 and the base angles equal to 30 . (7) Let ABCDEF be a cyclic hexagon with AB = CD = EF . Prove that the intersections of AC with BD, of CE with DF , and of EA with FB form a triangle similar to BDF . (8) Let ABC be a triangle, such that BC = a, AB = c, CA = b. Then show that: (a) If P is any arbitrary point, a.P B.P C + b.P C.P A + c.P A.P B ≥ abc If ABC is acute angled, and P is an interior point, the equality holds iff P is the orthocenter. (b) If P is any arbitrary point, a.P A2 + b.P B2 + c.P C2 ≥ abc (c) If P be an arbitrary point and G be the centroid, a.P A3 + b.P B3 + c.P C3 ≥ 3abc.P G ◦ (9) BC is a diameter of a circle with center O. A is a point on the circle with ∠AOC > 60 . EF is the chord which is the perpendicular bisector of AO. D is the midpoint of minor arc AB. The line through O parallel to AD meets AC again at J. Show that J is the incenter of the triangle CEF . (10) Let G be the centroid of the triangle ABC. Let R1,R2,R3 be circumradii of GBC, GCA, and GAB. Let R be the circumradius of ABC. Then:

R1 + R2 + R3 ≥ 3R Always Remember. Choose origin, axis, scaling carefully. Use linear transformations to simplify problems as far as one can. Of course, not all problems may have a convenient solution via complex numbers!

Vaibhav Vaish, The Institute of Mathematical Sciences, Chennai – 600113, India. Email Address: [email protected] 2