HOW TO FIND CUBE ROOT OF A GIVEN NUMBER
Problem 1 :
The complex fifth roots of 32 cos 5š 3 + i sin 5š 3
Solution:
Let z = 32 cos 5š 3 + i sin 5š 3
The given number is in polar form.
z = 32 cos 2kš + 5š 3 + i sin 2kš + 5š 3 Taking fifth root on both sides , z 1 5 = 32 cos 2kš + 5š 3 + i sin 2kš + 5š 3 1 5 Using De Moivre's theorem , bringing the power inside z 1 5 = 32 1 5 cos 6kš + 5š 3 + i sin 6kš + 5š 3 = 2 cos š 3 ( 6k + 5 ) + i sin š 3 ( 6k + 5 ) Put k = 0 , 1 , 2 , 3 and 4 When k = 0 = 2 cos š 3 ( 6 ( 0 ) + 5 ) + i sin š 3 ( 6 ( 0 ) + 5 ) = 2 cos 5š 3 + i sin 5š 3 --- ( 1 ) When k = 1 = 2 cos š 3 ( 6 ( 1 ) + 5 ) + i sin š 3 ( 6 ( 1 ) + 5 ) = 2 cos 11š 3 + i sin 11š 3 --- ( 2 ) When k = 2 = 2 cos š 3 ( 6 ( 2 ) + 5 ) + i sin š 3 ( 6 ( 2 ) + 5 ) = 2 cos 17š 3 + i sin 17š 3 --- ( 3 ) When k = 3 = 2 cos š 3 ( 6 ( 3 ) + 5 ) + i sin š 3 ( 6 ( 3 ) + 5 ) = 2 cos 23š 3 + i sin 23š 3 --- ( 4 ) When k = 4 = 2 cos š 3 ( 6 ( 4 ) + 5 ) + i sin š 3 ( 6 ( 4 ) + 5 ) = 2 cos 29š 3 + i sin 29š 3 --- ( 5 ) So , the roots are 2 cos 5š 3 + i sin 5š 3 , 2 cos 11š 3 + i sin 11š 3 , 2 cos 17š 3 + i sin 17š 3 , 2 cos 23š 3 + i sin 23š 3 and 2 cos 29š 3 + i sin 29š 3
Problem 2 :
The complex fifth roots of 32.
Solution:
nth roots of z = r cis Īø
z 1 n = r 1 n cis š + 2kš n for k = 0 , 1 , 2 ,.. n - 1 z 1 5 = 32 1 5 cis 0Ā° + 2kš 5 = 2 cis 0Ā° + 2kš 5 Put k = 0 , 1 , 2 , 3 and 4 When k = 0 = 2 ( cis 0Ā° ) = 2 --- ( 1 ) When k = 1 = 2 cis 2 ( 1 ) š 5 = 2 cis 2š 5 --- ( 2 ) When k = 2 = 2 cis 2 ( 2 ) š 5 = 2 cis 4š 5 --- ( 3 ) When k = 3 = 2 cis 2 ( 3 ) š 5 = 2 cis 6š 5 --- ( 4 ) When k = 4 = 2 cis 2 ( 4 ) š 5 = 2 cis 8š 5 --- ( 5 ) So , roots are 2 , 2 cis 2š 5 , 2 cis 4š 5 , 2 cis 6š 5 and 2 cis 8š 5
Problem 3 :
The complex sixth roots of 64.
Solution:
z 1 n = r 1 n cis š + 2kš n for k = 0 , 1 , 2 ,.. n - 1 z 1 6 = 64 1 6 cis 0Ā° + 2kš 6 = 2 cis 0Ā° + 2kš 6 Put k = 0 , 1 , 2 , 3 , 4 and 5 When k = 0 = 2 ( cis 0Ā° ) = 2 --- ( 1 ) When k = 1 = 2 cis 2 ( 1 ) š 6 = 2 cis š 3 --- ( 2 ) When k = 2 = 2 cis 2 ( 2 ) š 6 = 2 cis 2š 3 --- ( 3 ) When k = 3 = 2 cis 2 ( 3 ) š 6 = 2 ( cis š ) --- ( 4 ) When k = 4 = 2 cis 2 ( 4 ) š 6 = 2 cis 4š 3 --- ( 5 ) When k = 5 = 2 cis 2 ( 5 ) š 6 = 2 cis 5š 3 --- ( 5 ) So , roots are 2 , 2 cis š 3 , 2 cis 2š 3 , 2 ( cis š ) , 2 cis 4š 3 and 2 cis 5š 3
Problem 4 :
The complex cube roots of 1.
Solution:
z3 = 1
Let z = x + iy
(x + iy)3 = 1
z = r(cis Īø)
= cis 0
= cis(0 + 2kĻ)
= cis(2kĻ)
z = 3 1 = 1 1 3 = ( cis 2kš ) 1 3 z = cis 2kš 3 Put k = 0 , 1 and 2 When k = 0 z = cis 0 = 1 When k = 1 z = cis 2 ( 1 ) š 3 = cis 2š 3 = - 1 2 + i 3 2 When k = 2 z = cis 2 ( 2 ) š 3 = cis 4š 3 = - 1 2 - i 3 2 So , roots are 1 , - 1 2 + i 3 2 and - 1 2 - i 3 2
Problem 5 :
The complex cube roots of i.
Solution:
r = 1 š = š 2 z = 3 1 cis š 2 + 2kš 3 Put k = 0 , 1 , and 2 When k = 0 = 3 1 cis š 6 = 3 2 + 1 2 i --- ( 1 ) When k = 1 = 3 1 cis š 2 + 2 ( 1 ) š 3 = 3 1 cis 5š 6 = - 3 2 + 1 2 i --- ( 2 ) When k = 2 = 3 1 cis š 2 + 2 ( 2 ) š 3 = 3 1 cis 3š 2 = - i --- ( 3 ) So , roots are 3 2 + 1 2 i , - 3 2 + 1 2 i and - i
Problem 6 :
The complex fourth roots of 1 + i
Solution:
r = 1 2 + 1 2 = 2 tan š = y x = 1 1 š = š 4 1 + i = 2 cis š 4 z = 2 1 4 cis š 4 + 2kš 1 4 = 2 1 4 cis š 4 + 2kš 4 Put k = 0 , 1 , 2 and 3 When k = 0 = 8 2 cis š 16 When k = 1 = 8 2 cis š 4 + 2 ( 1 ) š 4 = 8 2 cis 9š 16 = - 8 2 cos 7š 16 + 8 2 i sin 7š 16 When k = 2 = 8 2 cis š 4 + 2 ( 2 ) š 4 = 8 2 cis 17š 16 = - 8 2 cos š 16 - 8 2 i sin š 16 When k = 3 = 8 2 cis š 4 + 2 ( 3 ) š 4 = 8 2 cis 25š 16 = 8 2 cos 7š 16 - 8 2 i sin 7š 16 So , roots are 8 2 cis š 16 , - 8 2 cos 7š 16 + 8 2 i sin 7š 16 , - 8 2 cos š 16 - 8 2 i sin š 16 and 8 2 cos 7š 16 - 8 2 i sin 7š 16
Problem 7 :
The complex fifth roots of -1 + i
Solution:
r = ( - 1 ) 2 + 1 2 = 2 tanš = y x = 1 - 1 = - 1 š = 3š 4 z = 2 1 5 cis 3š 4 + 2kš 5 Put k = 0 , 1 , 2 , 3 and 4 When k = 0 = 10 2 cis 3š 4 + 2 ( 0 ) š 5 = 10 2 cis 3š 20 --- ( 1 ) When k = 1 = 10 2 cis 3š 4 + 2 ( 1 ) š 5 = 10 2 cis 11š 20 --- ( 2 ) When k = 2 = 10 2 cis 3š 4 + 2 ( 2 ) š 5 = 10 2 cis 19š 20 --- ( 3 ) When k = 3 = 10 2 cis 3š 4 + 2 ( 3 ) š 5 = 10 2 cis 27š 20 --- ( 4 ) When k = 4 = 10 2 cis 3š 4 + 2 ( 4 ) š 5 = 10 2 cis 7š 4 --- ( 5 ) So , roots are 10 2 cis 3š 20 , 10 2 cis 11š 20 , 10 2 cis 19š 20 , 10 2 cis 27š 20 and 10 2 cis 7š 4