SOLUTION: 4th root of negative one

Question 12743: 4th root of negative one
Answer by khwang(438) (Show Source): You can put this solution on YOUR website!
Since -1 = cos pi + i sin pi
= cos (pi + 2k pi) + i sin (pi + 2 k pi)
= cos (2k+1) pi + i sin (2k+1) pi for integer k

By De Moivre�s Theorem, if z = r(cos x + i sin x)
then = (cos nx + i sin nx)
and = (cos + i sin )
for integer n.
Hence, = cos ((2k+1)*pi /4) + i sin ( (2k+1)*pi /4) )
k= 0,1,2 or 3.
So, we have four 4th root of -1 as:
cos () + i sin (), ( when k = 0) called the primitive 4th root of -1.
cos () + i sin (), (when k = 1)
cos () + i sin (), (when k = 2)
cos () + i sin (), (when k = 3)
Note use pi/4 (radians) not degrees. (even we know it is 45 deg)

This is very basic questions of complex numbers. Try to think about it.
Kenny

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