Three Cube Roots of Unity & Four Fourth Roots of Unity Three Cube Roots & Four Fourth Roots of a Number
Exercise 4.4 1. Find the three cube roots of: 8, -8, 27, -27, and 64 (I) 8 Solution: et x be a cube root of 8 1
∴
x = 8 !
= (8)
!
1
= ((8) ) x = 8 x − 8 = " ( x) − (2) = " ( x − 2)( x + 2 x + 4) = " x
!
!
!
! !
!
!
2
x − 2 = "
+ 2x + 4 = " −2 ± (2) − 4(1)(4) x =
x
2
2
x = 2
2(1)
x = x = x = x =
−2 ± −12 2
−2 ± −(4)(!) 2
−2 ± 2 −! 2 2(−1 ± !i )
x = 2(
⇒
2
−1 + 2
!i
x = 2(
)
x = 2(
−1 −
x = 2 w
x = 2w
2 2
−1 ±
!i
!i
2
)
) w= w2
−1 +
=
!i
2
−1 −
!i
2
2
#ence, three cube roots of 8 are: $2, 2w, 2w %
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(II) -8 Solution: et x be a cube root of -8 1
∴
x =
!
−8 = (−8)
!
1
= ((−8) ) x = −8 x + 8 = " ( x) + (2) = " ( x + 2)( x − 2 x + 4) = " x
!
!
!
! !
!
!
2
− 2x + 4 = " −(−2) ± (−2) − 4(1)(4) x =
x + 2 = "
x
2
2
x = −2
2(1)
x = x = x = x =
2±
−12 2
2±
−(4)(!) 2
2±2
−!
2 2(1 ± !i ) 2
1 + !i ) x = 2( 2 x = −2(
−1 − 2
1 ± !i ) x = 2( 2
⇒
1 − !i ) x = 2( 2 !i
)
x = −2w2
x = −2(
−1 +
x = −2w
2
!i
) w= w
2
−1 +
=
!i
2
−1 −
!i
2
#ence, three cube roots of -8 are: $−2, −2w, −2w % 2
(III) 27 Solution: et x be a cube root of 27 1
∴
x =
!
27
= (27)
!
1
x
!
x
!
x
!
= ((27) ) = 27 − 27 = " !
!
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( x)! − (!)! ( x − !)( x 2
=" + ! x + *) = " + !x + * = " −! ± (!) − 4(1)(*) x =
x − ! = "
x
2
2
x = !
2(1)
x = x = x = x = x =
−! ± −27 2
−! ± −(*)(!) 2
−! ± ! −! 2 !(−1 ±
−!)
2 !(−1 ± !i )
⇒
2
x = !(
−1 +
!i
2
x = !(
)
x = !(
−1 −
x = !w
x = !w
2
−1 ±
!i
2
!i
)
)
2
2 #ence, three cube roots of 27 are: $!, !w,!w %
(I+) -27 Solution: et x be a cube root of -27 1
∴
x =
!
−27 = (−27)
!
1
= ((−27) ) x = −27 x + 27 = " ( x) + (!) = " ( x + !)( x − ! x + *) = " x
!
!
!
! !
!
!
2
x + ! = "
− !x + * = " −(−!) ± (−!) − 4(1)(*) x = x
2
2
x = −!
2(1)
x = x =
!±
−27 2
!±
−(*)(!) 2
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x = x = x =
!±!
−!
2 !(1 ±
−!)
2 !(1 ± !i ) 2
1 + !i ) x = !( 2 x = −!(
−1 −
1 ± !i ) x = !( 2
⇒
1 − !i ) x = !( 2 !i
2
)
x = −!w2
x = −!(
−1 +
!i
2
)
x = −!w
2 #ence, three cube roots of -27 are: $−!, −!w, −! w %
(+) 64 Solution: et x be a cube root of 64 1
∴
x =
!
64 = (64) ! 1
= ((64) ) x = 64 x − 64 = " ( x) − (4) = " ( x − 4)( x + 4 x + 16) = " x
!
!
!
! !
!
!
2
x − 4 = "
+ 4 x + 16 = " −4 ± (4) − 4(1)(16) x = x
2
2
x = 4
2(1)
x = x = x = x =
−4 ± −48 2
−4 ± −(16)(!) 2
−4 ± 4 −! 2 4(−1 ± !i )
x = 4(
2
−1 + 2
!i
)
x = 4w
⇒ x = 4(
x = 4(
−1 − 2
−1 ±
!i
2
!i
)
)
x = 4 w2 2
#ence, three cube roots of 64 are: $4, 4w, 4w %
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2. aluate: 2 8 (i) (1 + w − w ) (i) (
−1 + −! 2
)* + (
(ii) w28 + w2* + 1
− 1− − ! 2
)7
2 2 (iii) (1 + w − w )(1 − w + w )
() ( −1 +
−!) + (−1 − .
!).
2 8 (i) (1 + w − w ) Solution: (1 + w − w2 )8
/dd and subtract w2 (1 + w − w2 )8 0 (1 + w + w2 − w2 − w2 )8 2 2 8 0 (1 + w + w − 2 w )
0 (" − 2w2 )8
2
Q1+ w + w
=
"
16
0 26w ! 1. 0 26( w ) w 1. 0 2.6(1) w 0 26w
!
Qw
=1
(ii) w28 + w2* + 1 Solution: ! * ! * 2 28 2* w + w + 1 0 (w ) w + (w ) w + 1 * * 2 0 (1) w + (1) w + 1
0 w + w2 + 1 0 "
2
Q w+ w
+
1 = 1 + w + w2
=
"
2 2 (iii) (1 + w − w )(1 − w + w ) Solution: (1 + w − w2 )(1 − w + w2 )
/dd and subtract w2 in the 1st bracet, and add and subtract w in the 2nd bracet. 2 2 2 2 (1 + w − w2 )(1 − w + w2 ) 0 (1 + w + w − w − w )(1 + w − w − w + w ) 2 2 2 0 (1 + w + w − 2 w )(1 + w + w − 2 w) 2 0 (" − 2 w )(" − 2w ) 2 0 ( −2w )(−2w)
0 4 w! 0 4 (i) (
−1 + −! 2
)* + (
− 1− − ! 2
)7
Solution:
&athe'atics otes for First Year (lass 11)
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(
−1 + −! 2
)* + (
−1− −! 2
* 2 7 0 ( w) + ( w )
)7
!
0 (w )
!
! ! 0 (w )
+w +(w )
Qw=
1+
−
−
2
!
, w2
=
−1 − −! 2
14
!
4
2
w
! 4 2 0 (1) + (1) w
0 1 + w2 0 −w If (
−1 + −!
)7
2 hen solution: (
−1 + −! 2
+(
)7 + (
−1 − − ! 2
−1 − − ! 2
)7
7 0 ( w)
)7
+
( w2 ) 7
! 2 14 0 (w ) w + w ! 2 ! 4 2 0 (w ) w + (w ) w
0 w + w2 0 −1 () (−1 + −!). + (−1 − !). Solution: &ulti3l and diide b 52 ith both the bracets. (−1 +
−!) + (−1 − .
−1 + −!
0 (2 ×
!).
2
0 (2 × w)
.
). + (2 ×
+ (2 × w
2
)
−1 − − ! 2
).
.
0 !2w + !2w ! 2 ! ! 0 !2w w + !2( w ) w .
1"
2 0 !2(1) w + !2(1)w
0 !2w2 + !2w 2 0 !2( w + w) 0 !2(−1) 0 −!2 !. Sho that: ! ! 2 (i) x − y = ( x − y)( x − wy)( x − w y) Solution: ! ! 2 x − y = ( x − y)( x − wy)( x − w y) x ! − y! x
!
−y
!
= ( x − wxy − xy + wy )( x − w y) = ( x − w x y − wx y + w xy − x y + w 2
!
2
2
2
2
2
!
2
2
2
2
xy
+ wxy − w 2
!
!
y)
earran9in9 e hae,
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x ! − y! x
!
x
!
x ! x
!
−y −y −y −y
!
!
!
!
= ( x − w x y − wx y − x y + w xy + w xy + wxy − w = x − x y( w + w + 1) + xy ( w + w + w) − w y = x − x y( w + w + 1) + wxy ( w + w + 1) − w y = x −w y =x −y roed. !
2
2
2
!
2
2
!
2
2
!
!
!
!
2
!
2
2
!
2
2
2
2
2
2
!
!
!
!
!
y! )
!
! ! ! 2 2 (ii) x + y + z − ! xyz = ( x + y + z)( x + wy + w z)( x + w y + wz ) Solution: ! ! ! 2 2 x + y + z − ! xyz = ( x + y + z)( x + wy + w z)( x + w y + wz ) ae .#.S 2 2 0 ( x + y + z )( x + wy + w z)( x + w y + wz ) 2 2 ! 2 0 ( x + y + z)( x + w xy + wxz + wxy + w y
+
w2 yz + w2 xz + w4 yz + w! z 2 )
2 2 2 2 ! 2 0 ( x + y + z )( x + xy( w + w) + xz( w + w ) + yz( w + w) + w y 2 2 0 ( x + y + z )( x + xy(−1) + xz(−1) + yz(−1) + (1) y 2 2 0 ( x + y + z )( x − xy − xz − yz + y )
0 x
!
0 x
!
− x y − x z − xyz + xy + x y − xy − xyz− + y + z − ! xyz roed. 2
!
2
2
2
2
y2 z + y! + x2 z− xyz− xz2 − yz2
+
y2 z
!
2 4 8 (iii) (1 + w)(1 + w )(1 + w )(1 + w ) ;; 2n factors 0 1 Solution: (1 + w)(1 + w2 )(1 + w4 )(1 + w8 ) ;; 2n factors
(1 + w)(1 + w2 )(1 + w)(1 + w2 ) (1 + w2 + w + w! )(1 + w2 (1 + w2 + w + 1)(1 + w2 (" + 1)(" + 1) (1)(1) (1) 1
+ w+ w ) + w + 1) !
;;
2n factors
;;
2n factors
;;
2n factors
;; ;;
2n factors 2n factors
2n
roed.
4. If w is a root of x 2 + x + 1 = " , sho that its other root is w2 and 3roe that w! Solution: If w is the root of the e
2
+ x +1 = " ⇒
= 1.
( w 2 ) 2 + w2 + 1 = " /dd and subtract w2 . ( w2 ) 2 + w2 + w2 + 1 − w2 ( w2 ) 2 + 2 w2 + 1 − w2
="
="
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( w2 + 1)2 − w2
=" ( w + 1 + w)( w + 1 − w) = " Qa −b Fro' w2 ± w ± 1 = " w4 − w = " w( w! − 1) = "
/s,
!
w≠"
w
!
w
−1 = " =1
roed.
. roe that co'3le? cube roots of −1 are *
*
1 + −! 1 − − ! + 2 2
1 + !i 2
and
1 − !i 2
@ and hence 3roe that
= −2
Solution: et x be the cube root of
−1 , therefore
1
x =
−1 = (−1) x = −1 x + 1 = " ( x) + (1) = " ( x + 1)( x − x + 1) = " x + 1 = " x − x + 1 = " −(−1) ± 1 − 4 x = −1 x = !
!
! !
!
!
2
2
2(1)
x = x = x =
1±
−! 2
1 ± !i 2 1 + !i 2
x =
#ence, three cube roots of −1 are: $−1,
1 − !i 2
1 + !i 1 − !i , % 2 2
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*
*
1 + −! 1 − − ! →
2
−w =
1−
2
−!
−w = 2
2 #ence, e
1+
−! 2
*
( − w ) + ( − w ) = −2 2
ae .#.S *
( −w ) + ( − w) 2
−w − w 18
*
0
*
−( w ) − ( w !
6
!
)! 0
−(1) − (1) 6
!
0
−2
roed.
6. If w is a cube root of unit, for' an e
=1
/nser
7. Find four fourth roots of: 16, 81, and 62 (i) 16 Solution: et x be a fourth root of 16, therefore, x = 4 16 1
x = 16 4 1
( x) 4
4
= 16 4
= 16 x − 16 = " ( x ) − ( 4 ) = " ( x + 4)( x − 4) = " x + 4 = " x = −4 x = ± −4 x 4 4
2
2
2 2
2
2
Qa
2
−
b2
=
( a + b)( a − b)
2
−4= " x = 4 x = ± 4 x
2 2
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x = ±2 x = ± 4i x = ±2i #ence four fourth roots of 16 are: 2, −2, 2i , −2i
(ii) 81 Solution: et x be a fourth root of 81, therefore, x = 4 81 1
x = 814 4
= 81 1
( x) 4
4
= 81 x − 81 = " ( x ) − ( * ) = " ( x + *)( x − *) = " x + * = " x = −* x = −* x = *i x
4 4
2
2
2
2
2
2 2
Qa
2
−
b2
=
( a + b)( a − b)
−* = " x = * x = * x = ±! x
2 2
x = ±!i
#ence, four fourth roots of 81 are: !, −!,!i , − !i (iii) 62 Solution: et x be a fourth root of 62, therefore, x = 4 62 1
x = 62 4 4
( x) = 62 x − 62 = " ( x ) − ( 2) = " ( x + 2)( x − 2) = " Qa x + 2 = " x − 2 = " x = −2 x = 2 x = −2 x = 2 x = ±. x = 2i x = ±i 1
4
4
4
2
2
2
2
2
2
2
2
2
2
−
b2
=
( a + b)( a − b)
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#ence, four fourth roots of 62 are: , −, i , −i 8. Sole the folloin9 e
2
x
2
2 2
x = ±2
#ence, solution set 0 $2, −2, 2i, −2i% (ii) ! y − 24! y = " Solution: ! y . − 24! y = " .
! y ( y 4 − 81) = " ! y = "
− 81 = " y = " ( y ) − (*) = " ( y + *)( y − *) = " Qa y + * = " y − * = " y = −* y = * y = ± −* y = ± * y = ±!i y = ±! #ence, solution set 0 $", !, −!,!i , − !i% y
4
2
2
2
2
2
2
2
2
2
2
−
b2
=
( a + b)( a − b)
(iii) x ! + x2 + x + 1 = " Solution: x ! + x2 + x + 1 = " 2 x ( x + 1) + 1( x + 1) = " ( x + 1)( x 2 + 1) = " x + 1 = " x = −1
x 2 + 1 = "
= −1 x = ± −1 x = ± i x 2
Qi
2
= −1
2
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x = ±i #ence, solution set 0 $−1, i , −i%
(i) x . − x = " Solution: x . − x = " x( x 4 − 1) = "
−1 = " x = " ( x ) − (1) = " ( x + 1)( x − 1) = " x + 1 = " x − 1 = " x = −1 x = 1 x = ±i x = ±1 #ence, solution set 0 $",1, −1, i , −i% . x = "
x
4
2
2
2
2
2
2
2
2
2
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