Demoivres_theorem.pdf

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De Moivre’s Theorem If n is any integer, (cosθ + i sinθ)n = cos nθ + i sin nθ If n is any fraction, one of the values of (cosθ + i sinθ)n is cos nθ + i sin nθ.  nπ   nπ  − nθ + i sin − nθ  2   2 

(sinθ + i cosθ)n = cos

If x = cosθ + i sinθ, then x + xn +

1 1 = 2 cosθ, x − = 2i sinθ x x

1 1 = 2cos nθ, x n − n = 2i sin nθ. n x x

1. The nth roots of a complex number form a G.P. with common ratio cis

2π n

which is denoted by ω.

2. The points representing nth roots of a complex number in the Argand diagram are concyclic. 3. The points representing nth roots of a complex number in the Argand diagram form a regular polygon of n sides. 4. The points representing the cube roots of a complex number in the Argand diagram form an equilateral triangle. 5. The points representing the fourth roots of complex number in the Argand diagram form a square. 6. The nth roots of unity are 1, w, w2 ,..... wn-1 where w = cis

2π n

7. The sum of the nth roots of unity is zero (or) the sum of the nth roots of any complex number is zero. 8. The cube roots of unity are 1,ω,ω2 where ω = cis ω=

2π 2 4π , ω = cis or 3 3

−1+ i 3 2 −1− i 3 ,ω = . 2 2

1 + ω+ ω2 = 0. ω3 = 1 The product of the nth roots of unity is (-1)n-1 . The product of the nth roots of a complex number Z is Z(-1)n-1 . ω, ω2 are the roots of the equation x2 + x + 1 = 0

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Very Short Answer Questions

1.

If n is an integer then show that (1 + i ) + (1 − i ) = 2n +1 cos 2n

2n

nπ 2

Solution : Let 1 + i = r {cos θ + i sin θ }

( r cos θ )

2

+ ( r sin θ ) = 2 ⇒ r = 2

cos θ =

1 1 P.V of θ = π / 4 sin θ = 2 2

2

π π π π   ∴1 + i = 2 cos + i sin  similarly (1 − i ) = 2 cos − i sin  4 4 4 4  

(1 + i )

2n

+ (1 − i ) = 2n

( 2)

2n

π π  cos + i sin  + 4 4  2n

( 2)

2n

π  π cos − i sin  4 4 

2n

2nπ 2nπ 2nπ 2nπ   = 2 n cos + i sin cos − i sin  4 4 4 4  

= 2n +1 cos

2.

nπ 2

Find the values of the following

(

(i) 1 + i 3

)

3

Solution : π   1 + i 3 = 2 cos + i sin π / 3 {by mod-amplitude form} 3  

π   1 + 1 3 = 8 cos + isin π / 3 3  

(

)

3

= 8 {cos π + i sin π }

3

{∵ ( cosθ + i sin θ )

n

= cos nθ + i sin nθ

}

= 8 {−1 + 0} = − 8

(ii) (1 − i )

8

Solution

(1 − i )

8

8

8

  1 1   π   π 4 = 2 −i   =  2 cos 4 − i sin 4   = 2 {cos 2π − i sin 2π } 2       2

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www.sakshieducation.com (iii) (1 + i )

16

(1 + i )

16

16

Solution

  π π  =  2  cos + i sin   = 2 {cos 2π + i sin 2π } 4 4   

= 256 5

 3 i  3 i (iv)  +  − −    2 2 2 2    

3

5

 3 i  3 i +  − −  Solution.  2   2 2   2

5

π π  π π  cos + i sin  − cos − i sin  6 6  6 6  5

cos

5π 5π 5π + i sin − cos + 1 sin π / 6 6 6 6

2i sin

3.

5

5π 1 = 2i =i 6 2

( )

(

Find all values of 1 − i 3

)

1 3

1

(1 − i 3 )

1 3

  1 3   3 =  2  − i  2     2 1

  π π 3 =  2  cos − i sin   3 3      −π = 2 cos    3 1 3

  −π  + i sin    3

1

3  

 π  π  2k π −  2k π −    3  + isn 3 = 2 cos   3 3     1 3

= 3 2 cis ( 6k − 1)

π 9

     k = 0, 1, 2    

k = 0, 1, 2

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4.

Find all values of ( −i ) 6

Solution : -

( −i )

1 6

  −π = cos    2

  −π  + i sin    2

1

 6  

 2 kπ − π / 2  = cis   k = 0, 1, 2, 3, 4, 5 6   1

∴ ( −1) 6 = cis ( 4k − 1)

5.

π 12

k = 0, 1, 2, 3

Find all values of (1 + i )

2/ 3

1

(1 + i )

2/3

2   π π   3 =   2  cos + 1 sin    4 4      

  π  =  2  cos + i sin π / 2   2   

π   2k π + 2 = 2 cis  3   1 3

1

= 2 3 cis ( 4k + 1)

π 6

1 3

   k = 0, 1, 2   k = 0, 1, 2

1

6.

Find all the values of ( −16 ) 4 1

( −16) 4 = ( 24 ) 4 ( −1) 4 1

1

1  2 kπ + π = 2 ( cisπ ) 4 = 2cis  4 

= 2cis ( 2k + 1)

π 4

  k = 0, 1, 2, 3 

k = 0, 1, 2, 3

1

7.

Find all values of ( −32 ) 5 1

( −32) 5

( ) 25

1 5

1

1

( −1) 5 = 2{cos π + i sin π }5 www.sakshieducation.com

www.sakshieducation.com 8.

If 1, ω, ω 2 are the cube roots of units then prove that

1 1 1 + = 2 + ω 1 + 2ω 1 + ω

Solution : L.H.S

1 1 + 2 + ω 1 + 2ω

3 (1 + ω ) 1 + 2ω + 2 + ω = ( 2 + ω )(1 + 2ω ) 2 + 4ω + ω + 2ω 2 =

=

(

3 (1 + ω )

)

2 1 + ω 2 + 5ω

(

3 −ω 2

)

−2ω + 5ω

∵1 + ω = − ω 2

1 + ω2 = ω −3ω 2 = = −ω 3ω =−

9.

1

ω

2

=

1 1+ω

If 1, ω, ω 2 are the cube roots of unity then prove that

( 2 − ω ) ( 2 − ω 2 ) ( 2 − ω10 )( 2 − ω11 ) = 49 Solution : - ( 2 − ω ) ( 2 − ω 2 ) ( 2 − ω10 )( 2 − ω11 ) =

{( 2 − ω ) ( 2 − ω )} {( 2 − ω ) ( 2 − ω )} {∵ω = {4− 2 (ω + ω + ω )}{4 − 2 (ω + ω ) + ω } 2

2

=(

10.

2

3

10

2

=ω

ω11 = ω 2 }

3

4 + 2 + 1)( 4 + 2 + 1) = 49

If 1, ω, ω 2 are the cube roots of units then prove that

( x + y + z ) ( x + yω + zω ) ( x + yω 2 + 2ω ) = x3 + y 3 + z 3 − 3xyz Solution: -

( x + y + z ) { x + yω + zω 2 }{ x + yω 2 + zω} ( x + y + z ) { x 2 + xyω 2 + xzω + xyω + xyω + y 2ω 3 + yzω 2 + xzω 2 + yzω + z 2ω 3} www.sakshieducation.com

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( x + y + z ) { x2 + y 2 + z 2 + xy (ω 2 + ω ) + yz (ω + ω 2 ) + zx (ω + ω 2 )} ( x + y + z ) { x 2 + y 2 + z 2 − xy − yz − zx} x3 + y 3 + z 3 − 3xyz

 

11. i) If x = cisθ θ then find the value of  x 6 + Sol:

1  . x6 

x = e iθ

i)

x 6 = ei6 θ 1 = e−6iθ 6 x x6 +

1 = ei6θ + e −6 θi 6 x

= cos 6θ + i sin 6θ + cos 6θ − i sin 6θ = 2 cos 6θ.

ii) If x = cisθ θ then find cube roots of 8. x = (8)1/ 3

x3 − 8 = 0 (x − 2)(x 2 + 2x + 4) =0 x = 2, x =

x=

−2 ± 4 − 16 2

−2 ± 2 3 i 2

x = −1 ± 3 i

Roots are 2, 2ω, 2ω2.

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www.sakshieducation.com 12 Prove that –ω ω and –ω ω2 are roots of z2 – z + 1 = 0 where ω and ω2 are the complex cube roots of unity. Sol:

z2 − z + 1 = 0

z=

1± 1− 4 2

z=

1 ± 3i 2

z=

−[−1 ± 3i] 2

z = −ω, −ω2

13. If 1, ω, ω2 are the cube roots of unity, then find the values of the following: i) (a + b)3 + (aω + bω2 ) 2 + (aω2 + bω)3 Sol:

i) (a + b)3 + (aω + bω2 ) 2 + (aω2 + bω)3 = a 3 + b3 + 3a 2 b + 3ab 2 + a 3ω3 + b3ω6 + 3a 2 ω2 ⋅ bω2 + 3aω⋅ b 2 ω4 + a 3ω6 + b3ω3 + 3a 2 bω4 ⋅ ω + 3b 2 ω2 ⋅ aω2 = a 3 + b3 + 3a 2 b(1 + ω + ω2 ) + a 3 + b3 + 3b 2 a(ω2 + ω + 1) + a 3 + b3 = 3(a 3 + b3 )

ii) (a + 2b) 2 + (aω2 + 2bω)2 + (aω + 2bω2 ) 2 Sol. (a + 2b) 2 + (aω2 + 2bω)2 + (aω + 2bω2 ) 2 = a 2 + 4b 2 + 4ab + a 2 ω4 + 4b 2 ω2 + 4abω3 + a 2 ω2 + 4b 2 ω4 + 4abω3 = a 2 (1 + ω + ω2 ) + 4b 2 (1 + ω2 + ω) + 4ab(1 + ω3 + ω2 ) = 12ab.

iii) (1 − ω + ω2 )3 2 3 Sol. (1 − ω + ω )

Now 1 + ω + ω2 = 0 1+ ω2 = −ω

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www.sakshieducation.com = (−ω − ω)3 = (−2)3 ω3 = −8

iv) (1 − ω)(1 − ω2 )(1 − ω4 )(1 − ω8 ) Sol. = (1 − ω − ω2 + ω3 )(1 − ω)(1 − ω2 ) = (1 − ω − ω2 + ω3 )(1 − ω − ω2 + ω3 ) = (1 + 1 + 1)(1 + 1 + 1) =9

 a + bω + cω2   a + bω + cω2  + 2   2   c + aω + bω   b + cω + aω 

v) 

 a + bω + cω2   a + bω + cω2   +  c + aω + bω2   b + cω + aω2   Sol. =

ω2 (a + bω + cω2 ) aω2 + bω3 + cω4 + cω2 + aω3 + bω4 ω2 (b + cω + aω2 )

1 ω2 ω = ω2 + 3 ω = ω2 +

⇒ ω2 + ω = −1

vi) (1 − ω)3 + (1 + ω2 )3 3 2 3 Sol. (1 − ω) + (1 + ω )

= (−ω2 )3 + (−ω)3 = −1 + (−1) = −2.

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www.sakshieducation.com  a + bω + cω2   a + bω + cω2  + 2   2   c + aω + bω   b + cω + aω 

vii) (1 − ω + ω2 )5 + (1 + ω − ω2 )5  (1 − ω + ω2 )5 + (1 + ω − ω2 )5

1 + ω2 = −ω = (−2ω)5 + (−2ω2 )5 = (−2)5 (ω2 + ω) = (−2)5 (−1) = 32 .

Short Answer Questions

1.

α , β are the roots of the equation x2 − 2 x + 4 = 0 then for any n ∈ N show that

α n + β n = 2n +1 cos

nπ 3

Solution: x2 − 2x + 4 = 0 ⇒ x =  

α = 2 cos

π 3

+ i sin

2 ± 4 − 16 2 ± 2i 3 = 2 2

π

π π   β = 2 cos − i sin  3 3 3  n

  π π    π π  α + β = 2  cos + i sin   + 2  cos − sin   3 3    3 3    n

n

n

 nπ nπ nπ nπ  = 2n cos + i sin + cos − i sin  3 3 3 3   nπ  nπ  n +1 = 2n  2 cos  = 2 cos 3  3 

2.

cos α + cos β + cos ϑ = 0 = sin α + sin β + sin ϑ = 0 then show that

(i) cos 3α + cos 3β + cos 3ϑ = 3 cos (α + β + ϑ ) (ii) sin 3α + sin 3β + sin 3ϑ = 3 sin (α + β + ϑ ) (iii) cos ( 2α − β − ϑ ) + cos {2 β − ϑ − α } + sin ( 2ϑ − α − β ) = 3 (iv) sin ( 2α − β − ϑ ) + sin ( 2 β − ϑ − α ) + sin ( 2ϑ − α − β ) = 0

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www.sakshieducation.com (v) cos 2α + cos 2β + cos 2ϑ = 0 (vi) sin 2α + sin 2β + sin 2ϑ = 0 (vii) cos2 α + cos 2 β + cos2 ϑ = 0 (viii) sin 2 α + sin 2 β + sin 2 ϑ = 3/ 2 (ix) cos (α + β ) + cos ( β + ϑ ) + cos (ϑ + α ) = 0 (x) sin (α + β ) + sin ( β + ϑ ) + sin (ϑ + α ) = 0 Solution : Let x = cos α + i sin α

y = cos β + i sin β : z = cos ϑ + i sin ϑ

x + y + z = ( cos α + cos β + cos ϑ ) + i ( sin α + sin β + sin ϑ )

x + y + z = 0 ⇒ x3 + y 3 + z 3 = 3xyz

Proof of (i) & (ii)

( cos α + i sin α )

3

+ ( cos β + i sin β ) + ( cos ϑ + i sin ϑ ) = 3cisα cisβ cisϑ 3

3

cis3α + cis3β + cis3ϑ = 3cis (α + β + ϑ )

( cos 3α + i sin 3α ) + ( cos 3β + i sin 3β ) + ( cos 3ϑ + i sin 3ϑ ) = 3cos (α + β + ϑ ) + 3i sin (α + β + ϑ ) By comparing real and imaginary parts on both sides cos 3α + cos 3β + cos 3ϑ + 3cos (α + β + ϑ ) sin 3α + sin 3β + sin 3ϑ = 3sin (α + β + ϑ )

Proof of (iii) & (iv) We know that sin 3α + sin 3β + sin 3ϑ = 3sin (α + β + ϑ ) x3 + y3 + z 3 x2 y2 z2 = 3⇒ + + =3 xyz yz zx xy cis 2α cis 2 β cis 2ϑ + + =3 cis β cisϑ cisϑ. cisα cisα cis β

cis ( 2α − β − ϑ ) + cis ( 2 β − ϑ − α ) + cos ( 2ϑ − α − β ) = 3

{cos ( 2α − β − ϑ ) + i sin ( 2α − β − ϑ )} + cos ( 2β − ϑ − α ) + 1sin ( 2β − ϑ − α ) + cos ( 2ϑ − α − β ) + 1sin ( 2ϑ − α − β ) = 3

Comparing real and imaginary parts on both sides

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www.sakshieducation.com cos ( 2α − β − ϑ ) + cos ( 2 β − ϑ − α ) + cos ( 2ϑ − α − β ) =3 sin ( 2α − β − ϑ ) + sin ( 2 β − ϑ − α ) + sin ( 2ϑ − α − β ) = 0

Proof of V & VI We know that x + y + z = 0 ∴

1 1 1 1 1 1 + + = + + x y z cos α + 1sin α cos β + i sin β cos ϑ + i sin ϑ

= cos α − i sin α + cos β − i sin β + cos ϑ − i sin ϑ 1 1 1 + + =0 x y z

x + y + z = 0 ⇒ ( x + y + z ) = 0 ⇒ x 2 + y 2 + z 2 + 2 xy + 2 y + 2 zx = 0 2

1 1 1  x 2 + y 2 + z 2 + 2 xyz  + + = 0 z x y

( cisα )

2

+ ( cisβ ) + ( cisϑ ) + 2 ( cisα cisβ cisϑ ) ( 0 ) 2

2

 1 1 1  ∵ + + = 0   x y z  cis 2α + cis 2 β + cis 2ϑ = 0 ⇒ ( cos 2α + cos 2 β + cos 2ϑ ) + i ( sin 2α + sin 2 β + sin 2ϑ ) = 0

By comparing real and imaginary parts on both sides cos 2α + cos 2 β + cos 2ϑ = 0 sin 2α + sin 2 β + sin 2ϑ = 0

Proof of (vii) From the above problem we can prove cos 2α + cos 2 β + cos 2ϑ = 0

2cos2 α + 2cos2 β − 1 + 2cos2 ϑ − 1 =0 2 {cos 2 α + cos 2 β + cos 2 ϑ} = 3 ∴ cos 2 α + cos 2 β + cos 2 ϑ =

3 2

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www.sakshieducation.com Proof viii cos 2α + cos 2 β + cos 2ϑ = 0

1 − 2sin 2 α + 1 −2 sin 2 β + 1 − 2sin 2 ϑ = 0

{

3 = 2 sin 2 α + sin 2 β + sin 2 ϑ ⇒ sin 2 α + sin 2 β + sin 2 ϑ =

}

3 2

Proof of (ix) and (x) In proving iv & v we proved 1 1 1 + + =0 x y z ∴ yz + zx + xy = 0 ∴ cisα cis β + cis β cisϑ + cisϑ cisα = 0

= cis (α + β ) + cos ( β + ϑ ) + cis (ϑ + α ) = 0

{cos (α + β ) + i sin (α + β )} + {cos ( β + ϑ )} + {cos (ϑ + α ) + i sin (ϑ + α )} = 0 By comparing real and imaginary parts on both sides cos (α + β ) + cos ( β + ϑ ) + cos (ϑ + α ) = 0 sin (α + β ) + sin ( β + ϑ ) + sin (ϑ + α ) = 0

3.

z 2n − 1 = i tan nθ If n is an integer and z = cisθ then show that 2 n z +1

Solution : z 2 n − 1 ( cos θ + i sin θ ) − 1 = z 2 n + 1 ( cos θ + i sin θ )2 n + 1 2n

=

=

=

cos 2nθ + i sin 2nθ − 1 cos 2nθ + i sin 2nθ + 1

− (1 −cos 2nθ ) + i sin 2nθ

(1 + cos 2nθ ) + i sin 2nθ

(

)

i 2 2 sin 2 nθ + 2i sin nθ cos nθ 2 cos nθ + 2i sin nθ cos nθ 2

{∵ − 1 = i } 2

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www.sakshieducation.com =

2 i sin nθ {cos nθ + i sin nθ }

{

2 cos nθ cos nθ + i sin nθ

}

= i tan nθ

4. If (1 + x)n = a 0 + a1x + a 2 x 2 + ... + a n x n , then show that i) a 0 − a 2 + a 4 − a 6 + ... = 2n / 2 cos ii) a1 − a 3 + a 5 − a 7 + ... = 2n / 2 sin

nπ 4

nπ . 4

Sol: (1 + x)n = a 0 + a1x + a 2 x 2 + ... + a n x n Put x = i (1 + i) n = a 0 + a1i + a 2i 2 + ... + a n i n n

  π π   2  cos 4 + i sin 4   = (a 0 − a 2 + a 4 ...) + i(a1 − a 3 + a 5 ...)   

nπ nπ   2n/2  cos + i sin  = (a 0 − a 2 + a 4 ...) + i(a1 − a 3 + a 5 ...) 4 4  

Equating Real parts both sides a 0 − a 2 + a 4 ... = 2n / 2 cos

nπ 4

Equating imaginary parts a1 − a 3 + a 5 ... = 2n / 2 sin

5.

nπ . 4

Solve the following equations (i) x4 − 1 = 0

Solution : (i) x4 − 1 = 0 ⇒ x4 = 1 1

1

∴ x = (1) 4 = ( cos 0 + i sin 0 ) 4 2kπ i sin 2kπ   = cos +  k = 0, 1, 2, 3 4 4  

= cos 0 + 1 sin 0 cis

π 2

cisπ cis

3π 2

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www.sakshieducation.com = 1, I, -1, -i = ± 1, ± i

(ii) x5 + 1 = 0 Solution : 1

x5 + 1 = 0 ⇒ ( cos π + i sin π ) 5  2k π + π x = cos  5 

∴ x = cis

π 5

, cis

  k = 0, 1, 2,3, 4 

3π 7π 9π cisπ , cis , cos 5 5 5

(iii) x9 − x5 + x 4 − 1 = 0 Solution : x9 − x5 + x 4 − 1 = 0

(

) (

)

x5 x 4 − 1 + 1 x 4 − 1 = 0

(x

4

)

(

)

− 1 = 0 : x5 + 1 = 0

Do (i) , (ii) to get the solution of (iii)

(iv) x4 + 1 = 0 Solution : 1

1

x 4 + 1 = 0 ⇒ x = ( −1) 4 = ( cisπ ) 4  2 kπ + π x = cis  4 

x = cis

π 4

, cis

  k = 0, 1, 2, 3 

3π 5π 7π cis cis 4 4 4

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www.sakshieducation.com 6.

If n is a positive integer then show that 1 q arc. tan  p n

1

( p + iq ) n + ( p − iq ) n = 2 ( p 2 + q 2 ) 2n cos  1

1

Solution : Let p + iq = r {cos θ + i sin θ } r cos θ = p ∴r =

p2 + q2 p

cos θ = tan θ =

r sin θ = q ⇒ r 2 = p 2 + q 2

p +q 2

2

sin θ =

q p + q2 2

 p q ⇒ θ = tan −1   p q 1

1

( p + iq ) n + ( p − q ) n = {r ( cos θ + i sin θ )} n + {r ( cos θ − i sin θ )} n 1

1

1  θ θ θ θ = r n cos + i sin + cos − i sin  n n n n  1

2 n 1   =  p 2 + q   2 cos θ  n    

(

= 2 p2 + q2

)

1 2n

1 q cos  tan −1  p n

π π    1 +sin 8 + i cos 8  Show that   1 + sin π − i cos π  8 8 

8/3

7.

= −1

Solution : π π  1 + sin 8 + i cos 8  LHS =   1 + sin π − i cos π  8 8 

8/3

8/ 3

 π  π  1 + cos  2 − π / 8  + i sin  2 − π / 8          1 +cos  π − π / 8  − 1 sin  π − π / 3    2  2  

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www.sakshieducation.com 3π 3π  1 + cos 8 + i sin 8   1 + cos 3π − i sin 3π 8 8 

8/3

    

3π 3π   2 3π  2 cos 16 + 2i sin 16 cos 16  =   2 cos 2 3π − 2i sin 3π cos 3π  16 16 16  

8/ 3

8

 3π  3π 3π   3 2cos cos + 1sin    16  16 16      2 cos 3π  cos 3π − i sin 3π      16  16 16     3π 3π  3π 3π   + i sin + i sin   cos  cos   16 16  16 16      3π 3π   3π     3π   cos 16 − i sin 16   cos  16 + i sin 16        2  3π 3π   + i sin   cos   16 16     2 3π 2 3π   cos 16 + sin 16   

8/ 3

8/ 3

3π 3π    cos 8 + i sin 8   

8/ 3

cos π + i sin π = − 1

8. Find the common roots of x12 – 1 = 0 and x4 + x2 + 1 = 0. Sol:

x12 − 1 = 0

x = (1)1/12

x = [ cos(2nπ) + i sin(2nπ) ]

1/12

n = 0,1, 2,...11 ...(1) x4 + x2 +1 = 0 (x 2 − 1)(x 4 + x 2 + 1) = 0 x6 −1 = 0 x = (1)1/ 6 x = [ cos(2nπ) + i sin(2nπ) ]

1/ 6

= cos

2nπ 2nπ + i sin , n = 0,1, 2,3, 4, 5...(2) 6 6

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www.sakshieducation.com Common roots to (1) and (2) π 2π 4π 5π cis , cis , cis , cis . 3 3 3 3

9. Find the number of 15th roots of unity, which are also 25th roots of unity. Sol:

x = (1)1/15

x = [cos 2nπ + i sin 2nπ]1/15 2nπ 2nπ + i sin 15 15 n = 0,1, 2,3,...14 x = cos

n = 3, m = 5 2π 2π + i sin 25 25 n = 9, m = 15 x = cos

cos

6π 6π + i sin 5 5

x = (1)1/ 25 x = [cos 2mπ + i sin 2mπ]1/ 25 x = cos

2mπ 2mπ + i sin 25 25

m = 0,1, 2, 3,...24 n = 6, m = 10 4π 4π + i sin 5 5 n = 12, m = 20 x = cos

cos

8π 8π + i sin 5 5

n = 0, m = 0 5 roots common.

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www.sakshieducation.com 10. If the cube roots of unity are 1, ω, ω2 then find the roots of the equation (x–1)3 + 8 = 0. Sol:

(x − 1)3 = −8

(x − 1) = (−8)3 x − 1 = −2 ⇒ x = −1 x − 1 = −2ω ⇒ x = −2ω + 1 x − 1 = −2ω2 ⇒ x = −2ω2 + 1

11. Find the product of all values of (1 + i)4/5. Sol:

(1 + i)4/5 i   1 = ( 2)4 / 5  +  2  2

= (2)

2/5

4/5

  π π   cos  2nπ + 4  + i sin  2nπ + 4       

4/5

n = 0,1, 2,3, 4 π 4 π 4  = (2)2 / 5 cos ⋅ + i sin ⋅  4 5 4 5 

...(1)

9π 4 9π 4   = (2)2 / 5 cos ⋅ + i sin ⋅  4 5 4 5 

...(2)

Product = (2 =2

2

2/5 5

) e

π  i  (1+9 +17 + 25+33)  5 

π i (85) e5

= 22 [cos17 π + i sin17 π] = −4

12. If z2 + z + 1 = 0, where z is a complex number, prove that 2

2

2

2

2

2

1  2 1   3 1   4 1   5 1   6 1    z + z  +  z + 2  +  z + 3  +  z + 4  +  z + 5  +  z + 6  = 12 z   z   z   z   z    

Sol:

let z = ω then

L.H.S. =

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www.sakshieducation.com 2

2

2

2

2

1  2 1   3 1   4 1   5 1   6 1   ω+ ω + ω + 2  + ω + 3  +ω + 4  + ω + 5  + ω + 6  ω   ω   ω   ω   ω    

2

= (−1)2 + (−1)2 + (2)2 + (−1)2 + (−1) 2 + (2) 2 = 12

Long Answer Questions

1. 1, α, α 2 , α3 ,...α n −1 be the nth roots of unity then prove that 0 if p ≠ kn 1p + α p + (α 2 ) p + (α 3 ) p + ... + (α n −1 )p =  Where p, k ∈ N. n if p = kn

Sol:

Now x n − 1 = 0 ⇒ x = (1)1/ n x = [cos 2nπ + i sin 2nπ]1/ n 2mπ 2mπ + i sin n n 2mπ 2mπ α = cos + i sin n n 2mpπ 2mpπ α p = cos + i sin n n x = cos

Now p = kn 1 + 1 + 1 + … n terms = n If p ≠kn value 1p + α p + (α 2 )p + (α3 ) p + ... + (α n −1 )p = 0 .

2. Prove the sum of 99th powers of the roots of the equation x7 – 1 = 0 is zero and hence deduce the roots of Sol:

x 6 + x 5 + x 4 + x 3 + x 2 + x +1 = 0 .

x 7 − 1 = 0 ⇒ x = (1)1/ 7 x = (cos 2kπ + i sin 2kπ)1/ 7 x = cos

2kπ 2kπ + i sin 7 7

k = 0,1, 2,3, 4, 5, 6

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www.sakshieducation.com x1 = 1, x 2 =

2π i e 7 ,x

3

=

4π i e 7 ,..., x

6

=

12 π i e 7

99 x199 + x 99 2 + x 3 + ...

2π

199 + e 7

⋅99i

4π

+e 7

⋅99i

+ ...e

12 π ⋅99i 7

=0

Roots of x 6 + x 5 + x 4 + x 3 + x 2 + x +1 = 0 be cos

2kπ 2kπ + i sin ; k = 1, 2, 3, 4, 5, 6. 7 7

∵ (x 7 ) − 1 = (x − 1)(x 6 + x 5 + x 4 + x 3 + x 2 + x +1) = 0

x = 1 is one root ⇒ cis

2kπ ; k = 1, 2, 3, 4, 5, 6 are roots of 7

x 6 + x 5 + x 4 + x 3 + x 2 + x +1 = 0 .

3. Solve (x – 1)n = xn, n is a positive integer. n

 x −1   x  =1  

Sol:

x −1 1/ n = (1) x

x −1 1/ n = [ cos 2mπ + i sin 2mπ] x x −1 2mπ 2mπ = cos + i sin x n n 1−

i 1 =e x

2m π n

2mπ 2mπ 1 − i sin = n n x mπ mπ mπ 1 2 sin 2 − 2i sin cos = n n n x

1 − cos

2 sin

mπ  mπ mπ  1 sin − i cos =  n  n n  x

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www.sakshieducation.com x=

1 mπ  mπ mπ  2sin sin − i cos  n  n n 

mπ   mπ sin i cos +  1 n n  = X mπ  mπ mπ   mπ mπ  − + 2sin sin i cos sin i cos n  n n   n n  mπ   mπ sin n + i cos n  = mπ  2 mπ mπ  2sin sin + cos 2  n  n n  mπ   mπ sin n + i cos n  = mπ 2sin n 1 mπ  ; m = 1, 2, 3, …(n – 1) 1 + i cot  2 n 

4. If m, n are integers and x = cosα α+ isinα α, y = cosβ β + isinβ β then prove that x m yn +

1 1 = cos(mα + nβ) and x m y n − m n = 2i sin(mα + nβ) . x y x y m n

Sol. xm = (cosα + isinα)m = cos mα + isin mα yn = (cosβ + isinβ)n = cos nβ + isin nβ ∴xmyn = (cos mα + isinmα) (cosnβ + isinnβ) = cos (mα + nβ) + isin (mα + nβ)

... (1)

1 1 = cos(mα + nβ) + i sin(mα + nβ) x y m n

= cos(mα + nβ) − i sin(mα + nβ)

...(2)

By adding (1) and (2), we get x m yn +

1 = cos(mα + nβ) x y m n

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www.sakshieducation.com By subtracting (2) from (1), we get x m yn −

1 = 2i sin(mα + nβ) x y m n

n

n

5. If n is a positive integer, show that (1 + i) + (1 – i) =

n +2 2 2

 nπ  cos   .  4 

1   1 +i  2  2

Sol. (1 + i) = 2 

π π  = 2  cos + i sin  4 4 

1   1 (1 − i) = 2  −i  2  2 π π  = 2  cos − i sin  4 4  (1 + i) = n

( 2)

n

π π   cos + i sin  4 4 

n

nπ nπ   = 2n / 2  cos + i sin  4 4   (1 − i) = n

( 2)

n

π π   cos − i sin  4 4 

...(1)

n

nπ nπ   = 2n / 2  cos − i sin  4 4  

...(2)

By adding (1) and (2), we get nπ   (1 + i) n + (1 − i)n = 2n/2  2 cos =2 4  

n +2 2

 nπ  cos    4  θ

 nθ 

6. If n is an integer then show that (1 + cos θ + i sin θ)n + (1 + cos θ − i sin θ)n = 2n +1 cosn   cos   2  2  Sol. L.H.S. = (1 + cos θ + i sin θ)n + (1 + cos θ − i sin θ)n = n

θ θ θ  θ θ θ  =  2 cos 2 + 2i sin cos  +  2 cos 2 − 2i sin cos  2 2 2  2 2 2 

n

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www.sakshieducation.com n n  θ θ  θ θ   cos + i sin  +  cos − i sin   2 2  2 2   

= 2n cos n

θ 2

= 2n cos n

θ nθ nθ nθ nθ  + i sin + cos − i sin   cos 2 2 2 2 2 

= 2 n cos n

θ nθ   2 cos  2 2 

= 2n +1 cos n

θ nθ cos = R.H.S. 2 2

7. If cosα α +cosβ β + cosγγ = 0 = sinα α + sinβ β + sinγγ prove that cos2 α + cos2β + cos2 γ =

3 =sin2 α + sin2 β + sin2 γ. 2

Sol. (cos α + i sin α) + (cos β + i sin β) + (cos γ + i sin γ ) = (cos α + cos β + cos γ ) + i(sin α + sin β + sin γ ) = 0 + i0 (cos α + i sin α) + (cos β + i sin β) + (cos γ + i sin γ ) = 0 ...(1)

Let x = cisα, y = cisβ, z = cisγ then x + y + z = 0 by (1), then x2 + y2 + z2 = –2(xy + yz + zx) 1

1

1

= −2xyz  + +  x y z 



= −2xyz[cos α − i sin α + cos β − i sin β + cos γ − i sin γ ] = −2xyz[(cos α + cos β + cos γ ) − i(sin α + sin β + sin γ )] = −2xyz(0 − i0) = 0 ∴ x 2 + y2 + z2 = 0 ⇒ (cos α + i sin α) 2 + (cos β + i sin β)2 + (cos γ + i sin γ ) 2 = 0 ⇒ cos 2α + i sin 2α + cos 2β + i sin 2β + cos 2γ + i sin 2γ = 0 ⇒ (cos 2α + cos 2β + cos 2γ ) + i(sin 2α + sin 2β + sin 2γ ) = 0 ∴ cos 2α + cos 2β + cos 2γ = 0 2 cos 2 α − 1 + 2cos 2 β − 1 + 2 cos 2 γ − 1 = 0 2(cos 2 α + cos 2 β + cos 2 γ ) = 3

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www.sakshieducation.com ∴ cos 2 α + cos 2 β + cos 2 γ =

3 2

1 − sin 2 α + 1 − sin 2 β + 1 − sin 2 γ = ∴ sin 2 α + sin 2 β + sin 2 γ =

3 2

3 . 2

8. Find all the value of ( 3 + i)1/ 4 . Sol. The modulus amplitude form of  3 i 3 + i = 2  +  = 2(cos 30° + i sin 30°)  2 2

 

1/ 4

π

Hence ( 3 + i)1/ 4 =  2cis  6 

π  2kπ +   6 , k = 0, 1, 2, 3 = 21/ 4  cis    4

 12kπ + π  = 21/ 4 cis   , k = 0, 1, 2, 3  24  = 21/ 4 cis(12k + 1)

π , k = 0, 1, 2, 3 24

∴ All the values of are ( 3 + i)1/ 4 are 21/ 4 cis

π 1/ 4 13π , 2 cis , 24 24 25π 1/ 4 37 π 21/ 4 cis , 2 cis 24 24

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www.sakshieducation.com 9. If 1, ω, ω2 are the cube roots of unity, prove that i) (1 − ω + ω2 )6 + (1 − ω2 + ω)6 = 128 = (1 − ω + ω2 )7 + (1 + ω − ω2 )7 3

Sol:

We use 1+ ω + ω2 = 1 +

(−1 + i 3) (−1 − i 3)  2π  + = 0 and ω3 =  cis  = cis2π = 1 2 2 3  

i) (1 − ω + ω2 )6 + (1 − ω2 + ω)6 = (−ω − ω)6 + (−ω2 − ω2 )6 = 26 (ω6 + ω12 ) = 26 (2) = 128 (1 − ω + ω2 )7 + (1 + ω − ω2 )7 = (−ω − ω)7 + (−ω2 − ω2 )7 = (−2)7 (ω7 + ω14 ) = (−2)7 (θ + θ2 ) = (−128)(−1) = 128.

ii) (a + b)(aω + bω2 )(aω2 + bω) = a 3 + b3 . (a + b)(aω + bω2 )(aω2 + bω)

= (a + b)(a 2ω3 + abω4 + abω2 + b 2ω3 )

(

= (a + b) a 2 + ab(ω + ω2 ) + b 2

)

= (a + b)(a 2 − ab + b 2 ) = a 3 + b3

iii)x2 + 4x + 7 = 0 where x = ω – ω2 – 2. x = ω − ω2 − 2

(x + 2) = ω − ω2 ⇒ (x + 2) 2 = ω2 + ω4 − 2ω3 ⇒ x 2 + 4x + 4 = ω2 + ω − 2 = −1 − 2 = −3 ⇒ x 2 + 4x + 7 = 0.

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