SOLUTION: 4th root of negative one

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Question 12743: 4th root of negative one
Answer by khwang(438) About Me  (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 z%5En+ = r%5En (cos nx + i sin nx)
and z%5E%281%2Fn%29+ = r%5E%281%2Fn%29 (cos x%2Fn + i sin x%2Fn)
for integer n.
Hence, %28-1%29%5E%281%2F4%29 = 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 (pi%2F4) + i sin (pi%2F4), ( when k = 0) called the primitive 4th root of -1.
cos (3%2Api%2F4) + i sin (3%2Api%2F4), (when k = 1)
cos (5%2Api%2F4) + i sin (5%2Api%2F4), (when k = 2)
cos (7%2Api%2F4) + i sin (7%2Api%2F4), (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