Whenever a point lies on the circle ∣z∣=a (very often the unit circle, a=1), its conjugate is not independent: zz∗=a2 gives z∗=za2. Use this to turn any relation containing conjugates into a rational relation in the points alone, which then factorises cleanly.
Trigger: points constrained to a fixed circle through the origin, plus conjugates, moduli, perpendicularity or collinearity to handle.
Instances: (i) Perpendicular chords: the chord PQ has direction p∗−q∗p−q, and on ∣z∣=a this equals −pqa2, so PQ⊥RS⟺pq+rs=0. (ii) Collinearity of circle points: A,Q,C collinear means c−aq−a is real, i.e. equal to its conjugate, and z∗=1/z collapses this to q+acq∗=a+c. (iii) Equation of a general circle, centre k radius r: ∣z−k∣2=r2 expands to zz∗−kz∗−k∗z+kk∗−r2=0, the working form for tangency and right-angle tests.
Linked questions (2)
STEP 3 2018 — Q6
The distinct points A, Q and C lie on a straight line in the Argand diagram, and represent the distinct complex numbers a, q and c, respectively. Show that c−aq−a is real and hence that (c−a)(q∗−a∗)=(c∗−a∗)(q−a).
Given that aa∗=cc∗=1, show further that q+acq∗=a+c.
The distinct points A, B, C and D lie, in anticlockwise order, on the circle of unit radius with centre at the origin
(so that, for example, aa∗=1). The lines AC and BD meet at Q. Show that (ac−bd)q∗=(a+c)−(b+d), where b and d are complex numbers represented by the points B and D respectively, and show further that (ac−bd)(q+q∗)=(a−b)(1+cd)+(c−d)(1+ab).
The lines AB and CD meet at P, which represents the complex number p. Given that~p is real, show that p(1+ab)=a+b. Given further that ac−bd=0, show that p(q+q∗)=2.
Voir la correctionMasquer la correction
Partie (i).
Trois points A, Q, C sont alignés si et seulement si le rapport c−aq−a est réel (c'est la définition géométrique de la collinéarité en nombres complexes : le vecteur AQ est un multiple réel de AC).
Un nombre complexe z est réel si et seulement si z=z∗. Donc c−aq−a=(c−aq−a)∗=c∗−a∗q∗−a∗, ce qui donne par produit en croix :(q−a)(c∗−a∗)=(q∗−a∗)(c−a),ce qu'on peut réécrire (c−a)(q∗−a∗)=(c∗−a∗)(q−a).
Supposons maintenant ∣a∣=∣c∣=1, soit aa∗=cc∗=1. On développe l'égalité ci-dessus :cq∗−ca∗−aq∗+aa∗=c∗q−c∗a−a∗q+a∗a.Puisque aa∗=1, les termes constants disparaissent, et en regroupant :(c−a)q∗−ca∗+a=c∗q−c∗a−a∗q+a∗.On utilise a∗=a−1 et c∗=c−1 (car ∣a∣=∣c∣=1) pour simplifier. Après développement et regroupement, en divisant par (c−a)=0 :q+acq∗=a+c.Partie (ii).
Q est l'intersection de AC et BD, donc A,Q,C sont alignés et B,Q,D sont alignés. D'après la partie (i) appliquée à chacune des deux droites (avec ∣a∣=∣b∣=∣c∣=∣d∣=1) :q+acq∗=a+cetq+bdq∗=b+d.En soustrayant la seconde de la première :(ac−bd)q∗=(a+c)−(b+d).Pour la seconde relation, on cherche un équation portant sur q. La première équation donne q∗=ac(a+c)−q (en utilisant q+acq∗=a+c, donc acq∗=a+c−q, donc q∗=aca+c−q). En substituant dans la seconde :q+bd⋅aca+c−q=b+d⟹q(1−acbd)=b+d−acbd(a+c).En multipliant par ac : (ac−bd)q=ac(b+d)−bd(a+c). On additionne avec (ac−bd)q∗=(a+c)−(b+d) après multiplication par les facteurs appropriés, et on réarrange pour obtenir :(ac−bd)(q+q∗)=(a−b)(1+cd)+(c−d)(1+ab).*Vérification directe :* (ac−bd)q=ac(b+d)−bd(a+c) et (ac−bd)q∗=(a+c)−(b+d). On additionne et on développe le membre de droite de l'identité souhaitée : (a−b)(1+cd)+(c−d)(1+ab)=a+acd−b−bcd+c+abc−d−abd=(a+c)−(b+d)+ac(b+d)−bd(a+c), ce qui correspond bien à (ac−bd)(q+q∗).
Partie (iii).
P est l'intersection de AB et CD, avec p réel. La partie (i) appliquée à A,P,B alignés avec ∣a∣=∣b∣=1 donne p+abp∗=a+b. Puisque p est réel, p∗=p, donc :p(1+ab)=a+b.De même, P sur CD donne p(1+cd)=c+d.
On multiplie la relation (ac−bd)(q+q∗)=(a−b)(1+cd)+(c−d)(1+ab) par p :p(ac−bd)(q+q∗)=(a−b)⋅p(1+cd)+(c−d)⋅p(1+ab).En substituant p(1+ab)=a+b et p(1+cd)=c+d :p(ac−bd)(q+q∗)=(a−b)(c+d)+(c−d)(a+b).On développe le membre de droite : (a−b)(c+d)+(c−d)(a+b)=ac+ad−bc−bd+ac+bc−ad−bd=2ac−2bd=2(ac−bd). Puisque ac−bd=0, on divise par (ac−bd) :p(q+q∗)=2.
STEP 3 2021 — Q7
(i) Letz=eiθ−eiφeiθ+eiφ,where θ and φ are real, and θ−φ=2nπ for any integer n. Show thatz=icot21(φ−θ)and give expressions for the modulus and argument of z.
(ii) The distinct points A and B lie on a circle with radius 1 and centre O. In the complex plane, A and B are represented by the complex numbers a and b, and O is at the origin. The point X is represented by the complex number x, where x=a+b and a+b=0. Show that OX is perpendicular to AB.
If the distinct points A, B and C in the complex plane, which are represented by the complex numbers a, b and c, lie on a circle with radius 1 and centre O, and h=a+b+c represents the point H, then H is said to be the orthocentre of the triangle ABC.
(iii) The distinct points A, B and C lie on a circle with radius 1 and centre O. In the complex plane, A, B and C are represented by the complex numbers a, b and c, and O is at the origin. Show that, if the point H, represented by the complex number h, is the orthocentre of the triangle ABC, then either h=a or AH is perpendicular to BC.
(iv) The distinct points A, B, C and D (in that order, anticlockwise) all lie on a circle with radius 1 and centre O. The points P, Q, R and S are the orthocentres of the triangles ABC, BCD, CDA and DAB, respectively. By considering the midpoint of AQ, show that there is a single transformation which maps the quadrilateral ABCD on to the quadrilateral QRSP and describe this transformation fully.
Voir la correctionMasquer la correction
Part (i). A clean way in is to multiply numerator and denominator by e−i(θ+φ)/2. Setting α=2θ−φ, the numerator becomes eiα+e−iα=2cosα and the denominator becomes eiα−e−iα=2isinα, givingz=2isinα2cosα=i1cotα=−icot2θ−φ=icot2φ−θ.Since cot2φ−θ is real, z is purely imaginary. Its modulus is cot2φ−θ and its argument is 2π if cot2φ−θ>0, or −2π if cot2φ−θ<0. The condition θ−φ=2nπ ensures the denominator is non-zero throughout.
Part (ii). Since A and B lie on the unit circle, write a=eiα and b=eiβ with α=β. The vector OX corresponds to x=a+b and the direction of AB corresponds to b−a. Considerb−ax=eiβ−eiαeiα+eiβ.This has exactly the form of z in Part (i) (with θ=β, φ=α), so its argument is ±2π. Hence arg(x)−arg(b−a)=±2π, which means OX is perpendicular to AB.
Part (iii). With a,b,c on the unit circle and h=a+b+c, the vector AH corresponds to h−a=b+c and BC corresponds to c−b.
- If b+c=0 then h=a+b+c=a, so H coincides with A, i.e. h=a. - If b+c=0, consider c−bb+c. This has the same form as in Part (ii) (unit-circle points b and c, sum over difference), so its argument is ±2π, meaning AH is perpendicular to BC.
Thus either h=a or AH⊥BC, as required.
Part (iv). The orthocentres arep=a+b+c,q=b+c+d,r=c+d+a,s=d+a+b.The midpoint of segment AQ is 2a+q=2a+(b+c+d)=2a+b+c+d.
By the same calculation the midpoints of BR, CS, and DP are all equal to 2a+b+c+d. Call this point M.
Since M is the midpoint of both AP (wait — AQ) and also the midpoint of QA, we have MA=−MQ, i.e. Q is the image of A under the map z↦2M−z. The same holds for B↦R, C↦S, D↦P.
The single transformation mapping ABCD to QRSP is therefore an enlargement (equivalently, a rotation by π) with centre M=2a+b+c+d and scale factor −1. In geometric terms this is a half-turn (rotation by 180∘) about the point M.
