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Date May 2022 Marks available 2 Reference code 22M.1.AHL.TZ2.12
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 2
Command term Express Question number 12 Adapted from N/A

Question

In the following Argand diagram, the points Z1, O and Z2 are the vertices of triangle Z1OZ2 described anticlockwise.

The point Z1 represents the complex number z1=r1eiα, where r1>0. The point Z2 represents the complex number z2=r2eiθ, where r2>0.

Angles α, θ are measured anticlockwise from the positive direction of the real axis such that 0α, θ<2π and 0<α-θ<π.

In parts (c), (d) and (e), consider the case where Z1OZ2 is an equilateral triangle.

Let z1 and z2 be the distinct roots of the equation z2+az+b=0 where z and a, b.

Show that z1z2=r1r2eiα-θ where z2 is the complex conjugate of z2.

[2]
a.

Given that Rez1z2=0, show that Z1OZ2 is a right-angled triangle.

[2]
b.

Express z1 in terms of z2.

[2]
c.i.

Hence show that z12+z22=z1z2.

[4]
c.ii.

Use the result from part (c)(ii) to show that a2-3b=0.

[5]
d.

Consider the equation z2+az+12=0, where z and a.

Given that 0<α-θ<π, deduce that only one equilateral triangle Z1OZ2 can be formed from the point O and the roots of this equation.

[3]
e.

Markscheme

z2=r2e-iθ          (A1)

z1z2=r1eiαr2e-iθ           A1

z1z2=r1r2eiα-θ           AG


Note: Accept working in modulus-argument form

 

[2 marks]

a.

Rez1z2=r1r2cosα-θ  =0           A1

α-θ=arcos0  r1,r2>0

α-θ=π2  (as 0<α-θ<π)           A1

so Z1OZ2 is a right-angled triangle           AG

 

[2 marks]

b.

EITHER

z1z2=r1r2eiα-θ=eiπ3  (since r1=r2)            (M1)


OR

z1=r2eiθ+π3  =r2eiθeiπ3            (M1)


THEN

z1=z2eiπ3           A1

 

Note: Accept working in either modulus-argument form to obtain z1=z2cosπ3+isinπ3 or in Cartesian form to obtain z1=z212+32i.

 

[2 marks]

c.i.

substitutes z1=z2eiπ3 into z12+z22             M1

z12+z22=z22ei2π3+z22  =z22ei2π3+1             A1

 

EITHER

ei2π3+1=eiπ3             A1


OR

z22ei2π3+1=z22-12+32i+1

=z2212+32i             A1

 

THEN

z12+z22=z22eiπ3

=z2z2eiπ3  and  z2eiπ3=z1             A1

so z12+z22=z1z2             AG

 

Note: For candidates who work on the LHS and RHS separately to show equality, award M1A1 for z12+z22=z22ei2π3+z22  =z22ei2π3+1, A1 for z1z2=z22eiπ3 and A1 for ei2π3+1=eiπ3. Accept working in either modulus-argument form or in Cartesian form.

 

[4 marks]

c.ii.

METHOD 1

z1+z2=-a  and  z1z2=b              (A1)

a2=z12+z22+2z1z2             A1

a2=2z1z2+z1z2=3z1z2             A1

substitutes b=z1z2 into their expression             M1

a2=2b+b  OR  a2=3b             A1


Note: If z1+z2=-a is not clearly recognized, award maximum (A0)A1A1M1A0.

 

so a2-3b=0              AG

 

METHOD 2

z1+z2=-a  and  z1z2=b              (A1)

z1+z22=z12+z22+2z1z2             A1

z1+z22=2z1z2+z1z2=3z1z2             A1

substitutes b=z1z2 and z1+z2=-a into their expression              M1

a2=2b+b  OR  a2=3b             A1


Note: If z1+z2=-a is not clearly recognized, award maximum (A0)A1A1M1A0.


so a2-3b=0              AG

 

[5 marks]

d.

a2-3×12=0

a=±6  z2±6z+12=0             A1

for a=-6:

z1=3+3i, z2=3-3i  and  α-θ=-5π3  which does not satisfy 0<α-θ<π             R1

for a=6:

z1=-3-3i, z2=-3+3i  and  α-θ=π3             A1

so (for 0<α-θ<π), only one equilateral triangle can be formed from point O and the two roots of this equation             AG

 

[3 marks]

e.

Examiners report

The vast majority of candidates scored full marks in parts (a) and (b). If they did not, it was normally due to the lack of rigour in setting out of the answer to a "show that" question. Part (c) was, though, more often than not poorly done. Many candidates could not use the given condition (equilateral triangle) to find z1 in terms of z2. Part (d) was well answered by a rather high number of candidates.

Only a handful of students made good progress in (e), not even finding the possible values for a.

a.
[N/A]
b.
[N/A]
c.i.
[N/A]
c.ii.
[N/A]
d.
[N/A]
e.

Syllabus sections

Topic 1—Number and algebra » AHL 1.12—Complex numbers – Cartesian form and Argand diag
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