Question 12743
 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^n }}} = {{{r^n}}} (cos nx + i sin nx) 
 and {{{z^(1/n) }}} = {{{r^(1/n)}}} (cos {{{x/n}}} + i sin {{{x/n}}}) 
 for integer n.

 Hence, {{{(-1)^(1/4)}}}  = 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/4}}}) + i sin ({{{pi/4}}}), ( when k = 0) called the primitive 4th root of -1.
 cos ({{{3*pi/4}}}) + i sin ({{{3*pi/4}}}), (when k = 1)
 cos ({{{5*pi/4}}}) + i sin ({{{5*pi/4}}}), (when k = 2)
 cos ({{{7*pi/4}}}) + i sin ({{{7*pi/4}}}), (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