Complex Number
Roots of unity and Factorization
1.
If
1, ω, ω2
are the cube roots of unity, prove that :
(i)
(a + ω – ω2) (a – ω + ω2) = a2 + 3,
(ii)
(1 + iω – ω2) ( 1 – ω + iω2) = 2,
(iii) (a + b) (a + bω) (a + bω2) = a3 + b3,
(iv) (a + b + c) (a + bω + cω2) (a + bω2 + cω) = a3 + b3 + c3 – 3abc ,
2.
Let
ω be a complex cube root of unity and n
(i)
Prove that
1+ω +ω =0.
(ii)
Prove that
x2 + y2 + z2 – xy – yz – zx has a linear factor
n
(iii) Deduce that x2 + y2 + z2 – xy – yz – zx
3.
If
is a positive integer, but not a multiple of
3.
2n
x + ωy + ω2z .
is a factor of (x – y)n + (y – z)n + (z – x)n .
ω is one of the complex cube roots of unity, show that
(x + a + b) (x + ωa + ω2b) (x + ω2a + ωb) ≡ x3 – 3abx + a3 + b3 .
Hence solve the equation x3 – px + q = 0
4.
5.
6.
If
(1 + x)n = c0 + c1x + c2x2 + … + cnxn ,
(a)
By setting x = 1, ω , ω2
(b)
Show that :
by putting p = 3ab
and q = a3 + b3 .
n being a positive integer,
in turn and adding show that , c0 + c3 + c6 +…
(c 0 + c 3 + ...)2 + (c1 + c 4 + ...)2 + (c 2 + c 5 + ...)2 =
1
3
=
1⎛ n
nπ ⎞
⎜ 2 + 2 cos ⎟
3⎝
3 ⎠
(4 n + 2) .
Find the quadratic factors with real coefficients of
(i)
x8 – 4x4 + 16
(ii)
x6 + 8x3 + 64
Factorize x9 – 1 and prove that
1
4π ⎞⎛
1
8π ⎞
1
1
2π ⎞⎛
⎛
+ 1 = ⎜ x + − 2 cos ⎟⎜ x + − 2 cos ⎟⎜ x + − 2 cos ⎟ ,
3
x
x
9 ⎠⎝
x
9 ⎠⎝
x
9 ⎠
⎝
(i)
x3 +
(ii)
4π
3
π
2π
sin sin sin
=
,
9
8
9
9
π ⎞⎛
2π ⎞⎛ 2
⎛
2 4π ⎞
2
(iii) 64⎜ cos 2 θ − cos 2 ⎟⎜ cos 2 θ − cos 2
⎟⎜ cos θ − cos
⎟ = 4 cos 3θ − 1 .
9 ⎠⎝
9 ⎠⎝
9 ⎠
⎝
7.
(i)
Show that
(ii)
Express
n −1
kπ ⎤ ⎡
⎛ kπ ⎞⎤
⎡
x 2 n − 1 = ( x − 1)( x + 1) Π ⎢ x − cis ⎥ ⎢ x − cis⎜ −
⎟ ,
k =1⎣
⎝ n ⎠⎥⎦
n ⎦⎣
1
x 2n − 1
in partial fractions with real linear and quadratic denominators.
1
8.
Hence show that
9.
(
( 2z + 1) Π ⎧⎨4z 2 + 4z + csc 2
1
3
k =1
16
rπ
2n + 1
⎩
kπ ⎫
⎬ .
8 ⎭
)
3
kπ ⎫
⎧
16 cos16 θ − sin 16 θ = cos 2θ Π ⎨cos 2 2θ + cot 2 ⎬ .
k =1
8⎭
⎩
Show that all the non-zero roots of the equation (1 + x)2n+1 = (1 – x)2n+1
± i tan
,
are given by
where r has values 1, 2, … , n.
π
tan 2
By putting n = 2 or otherwise, show that
10.
( z + 1)8 − z 8 =
Solve the equation (z + 1)8 – z8 = 0 , and prove that
5
× tan 2
2π
5
=5 .
Show that every root of the equation (z + 1)2n + (z – 1)2n = 0 , where n is a positive integer, is purely
imaginary.
If the roots are represented in the Argand diagram by points P1 , P2 , … , P2n
11.
2
2
if
O is the origin, OP1 + OP2 + … +
If
n is a positive integer, prove that
n −1
rπ
⎛
⎞
x 2 n − 1 = ( x − 1)( x + 1) Π ⎜ x 2 − 2 x cos
+ 1⎟
r =1 ⎝
⎠
n
(ii)
n
2 rπ
⎛
⎞
x 2 n +1 − 1 = ( x − 1) Π ⎜ x 2 − 2 x cos
+ 1⎟
r =1⎝
2n + 1 ⎠
n −1
Use the last result to prove that
Prove that
n −1
rπ ⎞
⎛
x n − x −n = (x − x −1 ) Π ⎜ x + x −1 − 2 cos ⎟ .
r =1 ⎝
n⎠
Π sin
r =1
, prove that,
= 2n(2n – 1) .
(i)
Deduce from (i) that, if x ≠ 0 ,
12.
OP2n2
rπ
2n
=
n
2
n −1
.
n −1
2 kπ ⎞ ⎤
⎛
⎡
x 2 n − 2 x n cos nθ + 1 = Π ⎢ x 2 − 2 x cos⎜ θ +
⎟ +1 .
k =0
⎝
⎣
n ⎠ ⎥⎦
Deduce the following results :
(i)
n −1
kπ ⎞
⎛
sin nα = 2 n −1 Π sin ⎜ α +
⎟
k =0
⎝
n ⎠
(ii)
n −1
2 kπ ⎞ ⎤
⎡
⎛
cos nα − cos nβ = 2 n −1 Π ⎢cos α − cos⎜ β +
⎟ .
k =0 ⎣
⎝
n ⎠ ⎥⎦
(iii) From (i), deduce, by logarithmic differentiation,
cot nα =
(iv) Deduce from (iii) that
1 n −1 ⎛
kπ ⎞
Σ cot ⎜ α +
⎟,
k =0
⎝
n
n ⎠
csc 2 nθ =
kπ ⎞
⎛
Σ csc 2 ⎜ θ +
⎟,
⎝
n
n ⎠
1
n −1
2 k =0
α≠
θ≠
rπ
n
rπ
n
,
.
2
13.
(i)
If
n is a positive integer, prove that
n −1
2kπ ⎞
⎡
⎛
2⎤
x 2 n − 2 x n a n cos nθ + a 2 n = Π ⎢ x 2 − 2 xa cos⎜ θ +
⎟+a ⎥ .
k =0 ⎣
⎦
⎝
n ⎠
(ii)
By taking x = a = 1 and in turn, θ = 2α
and θ = 2α – π , show that the value of
rπ ⎞ n −1
rπ ⎞
⎛
⎛
Π sin 2 ⎜ α + ⎟ + Π cos 2 ⎜ α + ⎟
r =0
r
0
=
⎝
⎝
n⎠
n⎠
n −1
is 22 – 2n
when n is odd, and 23 – 2n sin2 nα
when n is even.
(iii) Using derivatives, deduce from (i) that
nx
x
2n
n −1
(x n − a n cos nθ)
− 2 x n a n cos nθ + a 2 n
=
n −1
∑
r =0
2 rπ ⎞
⎛
x − a cos⎜ θ +
⎟
⎝
n ⎠
.
2rπ ⎞
⎛
2
2
x − 2 xa cos⎜ θ +
⎟+a
⎝
n ⎠
(iv) By taking the derivatives again and substituting a particular value for
x , show that,
when cos nθ ≠ 1,
n −1
∑
r =0
(v)
In
1
2rπ ⎞
⎛
1 − cos⎜ θ +
⎟
⎝
n ⎠
(i) , take x = i ,
=
n2
1 − cos nθ
.
a = 1 and n even, deduce that, if k is a positive integer,
2π ⎞
( 2k − 1) π ⎞
π⎞ ⎛
k
⎛
⎛
2 2 k −1 cos θ cos⎜ θ + ⎟ cos⎜ θ +
⎟ = ( − 1) − cos 2kθ .
⎟ ... cos⎜ θ +
⎝
⎠
⎝
k⎠ ⎝
k ⎠
k
and deduce that
2π ⎞
( 2k − 1) π ⎞
π⎞ ⎛
k
⎛
⎛
2 2 k −1 sin θ sin ⎜ θ + ⎟ sin ⎜ θ +
⎟ = ( − 1) (1 − cos 2kθ ) .
⎟ ... sin ⎜ θ +
⎝
⎠
⎝
k⎠ ⎝
k ⎠
k
(vi) Prove that :
n −1
Π sin
r =0
14.
(4r + 1)π
4n
1
= 22
The n points A0 , A1 , A2 , … , An-1
−n
.
are the vertices of a regular polygon of n sides which is
inscribed in a circle center O , radius a . P is a point such that OP = x
and the angle POA0 = θ .
n −1
Prove that
Π PA r = x 2 n − 2 x n a n cos nθ + a 2 n
k =0
3