2a Notes Demovres Theorem
2a Notes Demovres Theorem
2a Notes Demovres Theorem
3
1. If n is an integer, (cos + isin)n = cos n + i sin n 2. Find the value of 1 i 3 .
2. If n is a rational number, then one of the values of
(cos + i sin )n is cos n + i sin n. 3 3
1 3 1 3
1+ i 3
3
3. If z = r cis = r cis (2k + ), then A: = 2 + i 8 + i
2 2 2 2
z =r1/n 1/n
cis 2kπ + θ
n where k = 0, 1, 2, ...., n - 1. = 8(cos600 + isin600)3
1 By applying De Moivre’s theorem for an integral
4. If x = cos + i sin then = cos + i sin .
x
index.
1 1
and i) x + = 2cosθ ii) x - = 2isinθ = 8[cos3(600) + isin3(600)]
x x
= 8(cos1800 + isin1800) = 8 [-1 + i(0)] = - 8.
1
5. If x = cos + i sin then = cos - i sin and
x 3. Find the value of (1 - i)8. AIMS
n 1 8
i) x + = 2cosnθ 1 1
xn A: (1 - i)8 = 2 -i
2 2
n 1
ii) x - = 2isin nθ .
2
8
xn = (cos 450 - i sin 450)8
6. Cube roots of unity :
The roots of x3 = 1 are called cube roots of unity By applying De Moivre’s theorem for an integral
which are 1, , 2 where index.
-1 + i 3 -1 - i 3 = 24 [cos8(450) - isin8 (450)]
= , 2 =
2 2
= 24 [cos3600 - isin3600] = 16 [1 - i(0)] = 16
7. cis1.cis2 = cis(1 + 2)
cisθ1
= cis θ1 - θ2
(
8. 5 5
cisθ2 3 i 3 i
4. Find the value of .
9. cis1.cis2 .cis3..............cisn 2 2 2 2
= cis (1+ 2+ 3+............+n).
5 5
3 i 3 i
A: + - -
LEVEL - I (VSAQ) 2 2 2 2
1. Find the value of (1 + i)16. = (cos 300 + isin 300)5 - (cos 300 - isin 300)5
16
1 1 = cos5(300) + isin5(300) - [cos5(300) - isin5(300)]
A: (1 + i) = 2
16 +i =
2 2
cos150 0 + isin150 0 - cos150 0 - isin150 0
2
16
= cos45 0 + isin45 0 16
= 28 [cos16(450) - isin16 (450)] 1
= 2i sin1500 = 2i i.
= 256 [cos720 + isin720 ] = 256 [1 - i.0] = 256
0 0 2
1 De Moivre’s Theorem
MATHEMATICS - IIA AIMSTUTOTIAL
5. If A, B, C are the angles of a triangle such that 10.Solve x4 - 1 = 0.
x = cis A, y = cis B, z = cis C, then find xyz.
A: x4 - 1 = 0 (x2 + 1) (x2 - 1) = 0
A: Given that x = cis A, y = cis B, z = cis C
x2 + 1 = 0 or x2 - 1 = 0
Now xyz = cis A.cis B.cis C = cis(A+B+C)
x2 = - 1 or x2 = 1
= cos (A + B + C) + i sin (A + B + C)
x = 1 or x = 1
= cos 1800 + i sin 1800 = - 1 + i(0) = - 1.
x = ± i or x = ± 1.
6 1
6. If x = cis , then find the value of x 6 .
x 11. If the cube roots of unity are 1, , 2, then find
A: Given that x = cos + i sin . the roots of the equation (x - 1)3 + 8 = 0.
x6 = (cos + i sin ) = cos 6 + i sin 6. (x - 1)3 = -8 = (-2)3 = - 2 (1)1/3 = - 2 (1, , 2)
1 1 x - 1 = - 2, - 2, - 22
Now, 6 = = cos6 - i sin6.
x cos6θ + isin6θ x = 1 - 2, 1 - 2, 1 - 22
6 1 = - 1, 1 - 2, 1 - 2
Hence, x + 6 =
x
= cos6θ + isin6θ + cos6θ - isin6θ
LEVEL - I (LAQ)
AIMS
= 2cos 6.
1 If cos + cos + cos = 0 = sin + sin + sin,
7. Find the cube roots of 8. then show that
A: Let x = 3
8 ⇒ x3 = 8 i) cos 3 + cos 3 + cos 3= 3cos (+ + )
ii) sin 3 + sin 3 + sin 3=3sin (++ )
x3 = 23 = (2.1)3
A: Given: cos + cos + cos = 0 = sin + sin + sin
x = 2(11/3)
Let a = cos +isin,
= 2(1, , 2) b = cos + isin
= 2, 2, 2 .
c = cos + isin
a + b + c = (cos +cos+cos) + i(sin+sin+sin)
8. If , are the roots of the equation x2 + x + 1 = = 0 + i (0)
0, then prove that + + -1 -1 =0.
a+b+c = 0
2
b b 4ac 1 1 4 1 3 i a3 + b3 + c3 = 3abc
x ω, ω2
2a 2 2 (cos + isin)3 + (cos + isin)3 + (cos + isin)3
A: 4 1 1
4
α β4 α1β1 ω4 ω2 . 2 ω ω2 1 0 = 3(cis) (cis) (cis)
ω ω By applying DeMoivre’s Theorem for an integral
index, we get
cos α i sin α 4
cos 3 + i sin 3 + cos 3 + i sin 3 + cos 3 + i sin 3
9. Simplify .
sinβ i cosβ 8 = 3cis( + + )
(cos 3 + cos 3 + cos 3) + i(sin 3 + sin 3 + sin 3)
c o sα + is in α 4 co s α + is in α 4
A: = = = 3[cos( + + ) + i sin(+ + )]
-i
8 8
2
s in β + ico s β i c o s β - is in β
Equating of the real and imaginary parts on
= 2 cos θ2 cos 2
(a+b+c)2 = 0 nθ
n+1 n .
a2 + b2 + c2 + 2(ab + bc + ca) = 0
a2 + b2 + c2 = -2(ab + bc + ca)
cis2 + cis2 + cis2 = -2abc c1 a1 b1 4. In n is a positive integer, show that
4
n+2
= -2abc[0-i(0)]
= 2 cos π +i sin π
= 0. 4 4
cos2+ i sin2+cos2+ i sin2+cos2+ i sin2 = 0 1-i = 2 cosπ -i sin π
4 4
(cos2+cos2+cos2)+i(sin2+sin2+sin2)=0+i(0)
(1+i)n + (1 - i)n
Equating the real parts, we get
cos2 + cos2 + cos2 = 0 ................(1) n n
n
3 = 2(sin2 + sin2 + sin2)
2 2 . 2cos n4
sin2 + sin2 + sin2 = 32.
cos n4
n +1
22 .
3. Prove that (1+cos + isin)n + (1+cos-isin)n
cos n4
n+2
2
cos n . 2
= 2 n+1
cos n
2 2
A: Now (1+cos + isin)n + (1+cos -isin)n 5. If , are the roots of the equation
A: Given equation is x2 - 2x +4 = 0
2
x = -b ± b - 4ac
2a
3 De Moivre’s Theorem
MATHEMATICS - IIA AIMSTUTOTIAL
n nπ nπ nπ nπ
2 ± 24 - 16 = 2 cos 2 + i . sin 2 + cos 2 - i. sin 2
2
2 ± 212 i cos- isin nπ
= 2n. 2 cos
1 2
2 ± 22 3 i cos+ isin
nπ
= 2n+1 cos .
1± 3 i 2
n n A: Given: Z = cis
2 21 i. 23 2 21 i. 23 = cos + isin.
2n
Z - 1 = (cos+ isin) - 1
2n
n n
2 cos 3 i.sin 3 2 cos 3 i.sin 3 Now 2n 2n
Z + 1 (cos- isin) + 1
cos n
n+1
2 i 2sin ncosn - 2sin n
2
3
i 2sin ncosn + 2cos2n
2n 2n
p + q p + q
2n 2n 1
2 cos 4 i.sin 4 2 cos 4 i.sin 4 n
+ p +q
2 2 p q
-i
Using the De Moivre’s Theorem for an integral index p + q p +q
2 2 2 2
2n cos2n i.sin2n 2n cos2n i.sin2n By applying DeMoivre’s theorem for a rational
4 4 4 4
index, we get one value as
4 De Moivre’s Theorem
MATHEMATICS - IIA AIMSTUTOTIAL
8
(p2+q2)1/2n [(cos +isin)1/n + (cos -isin)1/n]
= one value of cos 3 +isin 3 3
8 8
p q
Where cos , sin
p2+q2 p2+q2 = cos + isin
1
p 2 +q 2 2n cos isin cos isin
n n n n
= -1 + i(0)
= -1.
1
p 2
+q 2 2n 2cos 1
n
1
10. Solve the equation x9 - x5 + x4 - 1 = 0.
p 2 +q 2 2n
n
q
2cos 1 T an -1
p
A: Given equation is x9 - x5 + x4 -1 = 0.
x5 (x4 -1) +1(x4 -1) = 0
1
p 2 +q 2 2n
n
q
2cos 1 arcT an
p
(x5 +1) (x4 -1) = 0
Now x5 + 1 = 0
9. Show that one value of
x5 = -1
8
1 + sin π
8
+ i cos π
8
3 = cos + isin
1 + sin π - i cos π = -1.
8 8 = cis
AIMS
1 + sin + icos = cis(2k + ), k z
Consider 8 8
A: 1 + sin - icos = cis(2k + 1), k z
8 8
x = [cis(2k+1)]1/5
1 + cos - + isin -
2 8 2 8 = cis(2k+1) 5 where k = 0, 1, 2, 3, 4.
1 + cos - - isin -
2 8 2 8 = cis 5 , cis 35 , cis, cis 75 , cis 95
16
2 cos 3 cos 3 +isin 3
16 16
2 cos 3 cos 3 - isin 3 LEVEL - II (VSAQ)
16 16 16
5 De Moivre’s Theorem
MATHEMATICS - IIA AIMSTUTOTIAL
2. Prove that - and - 2 are the roots of
z2 - z + 1 = 0, where and 2 are the complex
LEVEL - II (LAQ)
cube roots of unity.
A: z2 - z + 1 = 0. 1. Find all the roots of x11 - x7 + x4 - 1 = 0.
A: Given equation is x11 - x7 + x4 - 1 = 0
b b2 4ac
z
2a
x7 x 4 1 1 x 4 1 0
x 1 x 1 0
4 7
1 4 11
2
1
Now x4 - 1 = 0.
2 1
x 4 1 cis0 cis 0 2kπ cis2kπ
2
1 3i
1
2kπ
2 x cis2kπ 4 cis , k 0, 1, 2, 3
4
1 3 i 1 3 i
, kπ
2 2 = cis , k 0, 1, 2, 3
2
1 3 i 1 3 i π 3π
, cis0, cis , cisπ, cis
2 2 2 2
Also x7 + 1 = 0
AIMS
2
, . x7 = - 1 = cis = cis( + 2k)
x = [cis (2k+1) ]1/7
3. If 1, , 2 are the cube roots of unity, find the
value of (1 - + 2)3 . cis 2k 1 , k 0, 1, 2, 3, 4, 5, 6
7
3
(1 - + 2)3 = 1
2
1 2 0 3 5 7 9 11 13
cis , cis , cis , cis , cis , cis , cis .
7 7 7 7 7 7 7
3
Hence the required roots of the given equation are
2
3 3 3 5
cis0, cis , cis, cis , cis , cis , cis ,
2 2 7 7 7
83 7 9 11 13
cis , cis , cis , cis
8 . 7 7 7 7
4. If 1, , 2 are the cube roots of unity, find the 2. If (1+x)n = a0 + a1x + a2x2 + ..............anxn, then
value of (1 - ) (1 - 2) (1 - 4)(1 - 8). show that
n
A: (1 - ) (1 - 2) (1 - 4)(1 - 8) nπ
i) a0 - a 2 + a 4 ............ = 2 2 cos .
= (1 - ) (1 - 2)(1 - ) (1 - 2) 4
n
= [(1 - ) (1 - 2)]2 nπ
ii) a1 - a3 + a5 ............ = 2 2 sin .
4
= [1 - - 2 + ]2
Now (1 + x)n = a0 + a1x + a2x2+........+anxn .
= [1- ( + 2) + 1]2
Put x = i, then
= [2 - (-1)]2
a0 a1i a2i2 ........ anin 1 i
n
= 32
= 9. n
2 cos i sin .
4 4
By applying De Moivre’s theorem for an integral
index
6 De Moivre’s Theorem
MATHEMATICS - IIA AIMSTUTOTIAL
n
n n Part1: Let n be a positive integer. We prove the
a0 a1i a2 a3i a4 .......... 22 cos isin
4 4 theorem by using the principle of mathematical
a0 a2 a4 ....... i a1 a3 a5 ....... induction.
n n
n n Let P(n) be the statement:
sin . 22 cos i2 2
4 4 (cos + isin)n = cosn + i sin n
Equating real and imaginary parts both sides.
n If n =1, LHS = (cos + isin)1
n
a0 - a2 + a4 ................ 22 cos = cos + isin
4
n RHS = cos1 + isin1
n
a1 - a3 + a5 ................ 2 2 sin . = cos + isin
4
LHS = RHS
3. If z2 + z + 1 = 0, where z is a complex number,
Thus P(1) is TRUE.
2 2 2
1 2 1 3 1 Assume that P(k) is true.
prove that z + + z + 2 + z + 3 +
z z z (cos + isin)k = cosk + i sink
Multiplying bothsides by cos + isin, we get
2 2 2
4 1 5 1 6 1 (cos + i sin)k+1 = (cosk + isink) (cos + isin)
z + 4 + z + 5 + z + 6 = 12 . AIMS
z z z = cosk cos + i sink cos + i cosk
Now z2 + z + 1 = 0.
sin+ i2 sink sin
1 1 4 1 3 i = cos(k + ) + i sin(k + )
z , 2
2 2
= cos (k + 1) + i sin(k+1) .
Taking z = , we get
2 2 2
P(k+1) is TRUE
1 2 1 3 1
z z z 2 z 3 By induction, P(n) is true for all positive integers n.
z z
2 2 2
i.e. (cos + i sin)n = cosn +isin n for all n z+.
4 1 5 1 6 1
z 4 z 5 z 6 Part 2: If n = 0, LHS = (cos +isin)0
z z z
=1
2 2 2
1 2 1 3 1 RHS = cos0 + isin0
= 2 3
=1
2 2 2 LHS = RHS
4 1 5 1 6 1
4 5 6 If n =1, the statement is TRUE.
2 2 2 2 Part 3: Let n be a negative integer and n = -m,
3 3 12 3 3 12
2 2 1 2 2 1 where m z+
1 1 So for m, part 1 is applicable.
2 2 2 2
2 2 2 2 2 2
2 2 Now (cos + isin)n = (cos + isin)-m
1
1 1 4 1 1 4
2 2 2 2
m
cos+ isin
12 .
1 from Part 1
cos m+ isin m
4. State and prove De Moivre’s Theorem for an
integral index. = cos m - isin m
A: De Moivre’s Theorem for an integral index: = cos(-m) + i sin(-m)
For any real number and any integer n, = cos n + i sin n.
(cos + isin)n = cosn + i sin n.
7 De Moivre’s Theorem
MATHEMATICS - IIA AIMSTUTOTIAL
5. If 1, , 2 are the cube roots of unity, prove
that
i) (1- +2)6 + (1-2+)6 = 128 = (1-+2)7 + (1+-2)7
ii) (a + b) (a+ b2) (a2 + b) = a3 + b3.
A: Given that 1, , 2 are the cube roots of unity,
then 1 + + 2 = 0 and 3 = 1
i) (1 - + 2)6 + (1 - + 2)6
= (-- )6 + (-2 - 2)6
= (-2)6 + (-22)6
= (-2)6 [6 + 12]
= 64(1 + 1)
= 128
(1 - + 2)7 + (1 + - 2)7
= (-- )7 + (-2 - 2)7
= (-2)7 + (-22)7
AIMS
= (-2)7 [7 + 14]
=(-128) ( + 2)
= (-128) (-1)
= 128.
ii) (a + b) (a+ b2) (a2 + b)
= (a + b) (a23 + ab 2 + ab4 + b2 3)
= (a + b) [a2 + ab(2 + ) + b2]
= (a + b) [a2 + ab (-1) + b2]
= (a + b) (a2 - ab + b2)
= a3 + b3
2 cos + i sin
4 4
2 cis
4
, kz
2 cis 2k 4
2 2
2 3 3
(1+ i) 2 cis 8k +1 4
3
1
= 2 cis 8k +1 . 2 . ,
3
k = 0, 1, 2
3 4
1
= 2 cis 8k +1 ,
3
k = 0, 1, 2
6
8 De Moivre’s Theorem