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Date May 2021 Marks available 3 Reference code 21M.3.AHL.TZ2.2
Level Additional Higher Level Paper Paper 3 Time zone Time zone 2
Command term Express Question number 2 Adapted from N/A

Question

This question asks you to investigate and prove a geometric property involving the roots of the equation zn=1 where z for integers n, where n2.


The roots of the equation zn=1 where z are 1, ω, ω2, , ωn-1, where ω=e2πin. Each root can be represented by a point P0, P1, P2, , Pn-1, respectively, on an Argand diagram.

For example, the roots of the equation z2=1 where z are 1 and ω. On an Argand diagram, the root 1 can be represented by a point P0 and the root ω can be represented by a point P1.

Consider the case where n=3.

The roots of the equation z3=1 where z are 1, ω and ω2. On the following Argand diagram, the points P0, P1 and P2 lie on a circle of radius 1 unit with centre O(0, 0).

Line segments [P0P1] and [P0P2] are added to the Argand diagram in part (a) and are shown on the following Argand diagram.

P0P1is the length of [P0P1] and P0P2 is the length of [P0P2].

Consider the case where n=4.

The roots of the equation z4=1 where z are 1, ω, ω2 and ω3.

On the following Argand diagram, the points P0, P1, P2 and P3 lie on a circle of radius 1 unit with centre O(0, 0). [P0P1], [P0P2] and [P0P3] are line segments.

For the case where n=5, the equation z5=1 where z has roots 1, ω, ω2, ω3 and ω4.

It can be shown that P0P1×P0P2×P0P3×P0P4=5.

Now consider the general case for integer values of n, where n2.

The roots of the equation zn=1 where z are 1, ω, ω2, , ωn-1. On an Argand diagram, these roots can be represented by the points P0, P1, P2, , Pn-1 respectively where [P0P1], [P0P2], , [P0Pn-1] are line segments. The roots lie on a circle of radius 1 unit with centre O(0, 0).

P0P1 can be expressed as |1-ω|.

Consider zn-1=(z-1)(zn-1+zn-2+  +z+1) where z.

Show that (ω-1)(ω2+ω+1)=ω3-1.

[2]
a.i.

Hence, deduce that ω2+ω+1=0.

[2]
a.ii.

Show that P0P1×P0P2=3.

[3]
b.

By factorizing z4-1, or otherwise, deduce that ω3+ω2+ω+1=0.

[2]
c.

Show that P0P1×P0P2×P0P3=4.

[4]
d.

Suggest a value for P0P1×P0P2×  ×P0Pn-1.

[1]
e.

Write down expressions for P0P2 and P0P3 in terms of ω.

[2]
f.i.

Hence, write down an expression for P0Pn-1 in terms of n and ω.

[1]
f.ii.

Express zn-1+ zn-2+  +z+1 as a product of linear factors over the set .

[3]
g.i.

Hence, using the part (g)(i) and part (f) results, or otherwise, prove your suggested result to part (e).

[4]
g.ii.

Markscheme

METHOD 1

attempts to expand (ω-1)(ω2+ω+1)            (M1)

=ω3+ω2+ω-ω2-ω-1           A1

=ω3-1           AG

 

METHOD 2

attempts polynomial division on ω3-1ω-1            M1

=ω2+ω+1           A1

so (ω-1)(ω2+ω+1)=ω3-1           AG

 

Note: In part (a), award marks as appropriate where ω has been converted into Cartesian, modulus-argument (polar) or Euler form.

 

[2 marks]

a.i.

(since ω is a root of z3=1)ω3-1=0           R1

and ω1           R1

ω2+ω+1=0           AG

 

Note: In part (a), award marks as appropriate where ω has been converted into Cartesian, modulus-argument (polar) or Euler form.

  

[2 marks]

a.ii.

METHOD 1

attempts to find either  P0P1 or P0P2             (M1)

accept any valid method

e.g.  2sinπ3,  12+12-2cos2π3,  1sinπ6=P0P1sin2π3from either ΔOP0P1 or ΔOP0P2

e.g. use of Pythagoras’ theorem

e.g. 1-ei2π3,  1--12+32i by calculating the distance between 2 points

P0P1=3           A1

P0P2=3           A1


Note:
Award a maximum of M1A1A0 for any decimal approximation seen in the calculation of either P0P1 or P0P2 or both.


so P0P1×P0P2=3            AG

 

METHOD 2

attempts to find P0P1×P0P2=1-ω1-ω2             (M1)

P0P1×P0P2=ω3-ω2-ω+1           A1

=1-ω2+ω+1+2  and since ω2+ω+1=0           R1

so P0P1×P0P2=3            AG

 

[3 marks]

b.

METHOD 1

z41=z1z3+z2+z+1           A1

(ω is a root hence) ω4-1=0 and ω1           R1

ω3+ω2+ω+1=0            AG


Note: Condone the use of ω throughout.

 

METHOD 2

considers the sum of roots of z4-1=0             (M1)

the sum of roots is zero (there is no z3 term)         A1

ω3+ω2+ω+1=0            AG

 

METHOD 3

substitutes for ω             (M1)

e.g. LHS=ei3π2+eπi+eiπ2+1

=-i-1+i+1         A1


Note: This can be demonstrated geometrically or by using vectors. Accept Cartesian or modulus-argument (polar) form.

ω3+ω2+ω+1=0            AG

 

METHOD 4

ω3+ω2+ω+1=ω4-1ω-1         A1

=0ω-1=0 as ω1           R1

ω3+ω2+ω+1=0            AG

 

[2 marks]

c.

METHOD 1

P0P2=2           A1

attempts to find either P0P1 or P0P3             (M1)

 

Note: For example P0P1=1-i and P0P3=1+i.
         Various geometric and trigonometric approaches can be used by candidates.

 

P0P1=2, P0P3=2           A1A1


Note: Award a maximum of A1M1A1A0 if labels such as P0P1 are not clearly shown.
         Award full marks if the lengths are shown on a clearly labelled diagram.
         Award a maximum of A1M1A1A0 for any decimal approximation seen in the calculation of either P0P1 or P0P3 or both.


P0P1×P0P2×P0P3=4            AG

 

METHOD 2

attempts to find P0P1×P0P2×P0P3=1-ω1-ω21-ω3            M1

P0P1×P0P2×P0P3=-ω6+ω5+ω4-ω2-ω+1         A1

=--1+ω5+1--1-ω+1  since ω6=ω2=-1 and ω4=1        A1

=ω5-ω+4 and since ω5=ω           R1

so P0P1×P0P2×P0P3=4            AG

 

METHOD 3

P0P2=2           A1

attempts to find P0P1×P0P3=1-ω1-ω3            M1

P0P1×P0P3=ω4-ω3-ω+1         A1

=2--ω-ω  since ω4=1 and ω3=-ω               R1     

so P0P1×P0P2×P0P3=4            AG

 

[4 marks]

d.

P0P1×P0P2×  ×P0Pn-1=n         A1

 

[1 mark]

e.

P0P2=1-ω2, P0P3=1-ω3         A1A1

 

[2 marks]

f.i.

P0Pn-1=1-ωn-1         A1A1

 
Note: Accept 1-ω from symmetry.


[1 mark]

f.ii.

zn-1=z-1zn-1+ zn-2+  +z+1

considers the equation zn-1+ zn-2+  +z+1=0         (M1)

the roots are ω, ω2,  , ωn-1         (A1)

so z-ωz-ω2z-ωn-1         A1


[3 marks]

g.i.

METHOD 1

substitutes z=1into z-ωz-ω2z-ωn-1zn-1+ zn-2+  +z+1           M1

1-ω1-ω21-ωn-1=n         (A1)

takes modulus of both sides           M1

1-ω1-ω21-ωn-1=n

1-ω1-ω21-ωn-1=n                 A1

so P0P1×P0P2××P0Pn-1=n                 AG

 

Note: Award a maximum of M1A1FTM1A0 from part (e).

 

METHOD 2

1-ω,1-ω2,,1-ωn-1 are the roots of 1-vn-1+1-vn-2++1-v+1=0           M1

coefficient of vn-1 is -1n-1 and the coefficient of 1 is n                 A1

product of the roots is -1n-1n-1n-1=n                 A1

1-ω1-ω21-ωn-1=n                 A1

so P0P1×P0P2××P0Pn-1=n                 AG


[4 marks]

g.ii.

Examiners report

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

Syllabus sections

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