Question 9:
9 | (a) | Diagram showing  as the ‘first’ non-real
1
vertex of a regular n-gon with 1 as the 0th
vertex and at least one other vertex shown
with the correct relationship (ie on unit circle
with same angular distance).
(Since it is a root of unity)the modulus of
1
is 1 so multiplying by it leaves the modulus
unchanged....
...(since the n roots of unity are represented
by the n vertices on the unit circle of a
regular n-gon then)rotation by the argument
of the first () (ie adding an angle) takes
1
you to the second and so on. | B1
B1
B1 | 2.1
2.2a
2.4 | Dealing with modulus (could be
incorporated in the below).
Dealing with argument. Accept
a well-reasoned argument based
on multiplication by 
1
representing a pure rotation (by
the required angle).
Could be argued by induction if
rigorous. | Diagram should clearly show equal
angular distance between the roots
and an equal distance (of 1) from O
to each root.
At least 3 points including  and
0
 shown.
1
For this B1 allow an n-gon with a
specific value of n chosen.
1 s i n c e i t i s a r o o t o f u n i t y  =
k
1 ( k1 k ) 1    =  = =
1 1
2 k 
a r g ( )  =
k n
2 
a r g ( )  =
1 n
a r g ( k1 ) k a r g ( ) 2 k     = =
1 n
Alternative method for last B1B1: | B1
2 2k
i i
=e n and  =e n
1 k | 𝜔 can be implied if appearing
1
in the equation below
B1
k
2n 2 k    
i i
k1 e e n    = = =
k | B1
[3]
(b) | n 1 n 1 − −
(0 = 1 = and so) k1    =  which is a
1 0 k
k 0 k 0 = =
GP with (a = 1), r =  ( 1) and n terms.
1
1(n −1) 1−1 0
= 1 = = =0(since  is
−1 −1 −1 1
1 1 1
an nth root of unity so n = 1).
1 | M1
A1
[2] | 3.1a
2.2a | Using the identity from (a) and
recognising the GP (can be
implied by the formula).
AG so reasoning must be
shown, | n 1 n 1 − −
k1    =  and recognition of
k
k 0 k 0 = =
n−1=(−1)(n−1+n−2+...++1)
1 1 1 1 1
If GP not recognised then
justification for  – 1  0 must
1
also be given.
(c) | z = a + b i
z * = a − b i
 z + z * = 2 a = 2 R e ( z ) | B1
[1] | 2.1 | AG
(d) | n−1 
 =0

k
 
k=0
n−1
* =0
k
k=0
n−1 n−1 n−1
 +* =0 or (  +* ) =0
k k
k k
k=0 k=0 k=0
n−1
2Re()=0
k
k=0
n−1
Re()=0
k
k=0 | B1
[1] | 3.1a
(e) | The roots of unity form a regular n-gon
which is symmetrical in the real axis. | B1
[1] | 2.1 | Or the (non-real) roots of unity
come in complex conjugate
pairs (since they are roots of the
real polynomial zn = 1). | Could use symmetry of cos
function in geometric context
2  2 
(eg cos =cos2−  etc)
5  5 
(f) | (i) | 2 
a r g  = soi
1 5
So from (d) and (e),
2 4  
1 2 c o s 2 c o s 0 + + =
5 5
4 2    
2 c o s 1 2 c o s  = − −
5 5
4 1 2  
c o s c o s  = − −
5 2 5
1
a , b 1 = − = −
2 | M1
A1
[2] | 3.1a
2.2a | Finding the first argument.
Could be embedded.
a and b can be embedded. | Could see sum of all 5 real parts.
(f) | (ii) | 2 1 2  
2 c o s 2 1 c o s  − = − −
5 2 5
2 
4 c 2 2 c 1 0 , c c o s  + − = =
5
− 1  5
c =
4
2  2
is an acute angle so cos 0.
5 5
− 1 − 5 2 1 5  − +
So is rejected so c o s . =
4 5 4 | M1
A1
[2] | 3.1a
2.3 | Using correct double angle
formula to derive and solve a
2 
quadratic equation in c o s
5
(must have real solutions).
Could be BC.
Clear rejection after valid
argument leading to correct
answer.
PMT
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