Question 2916
  You should type as : x^4 = -3
  Since -3 = 3*-1 = 3(cos pi + i sin pi)
  Solve x^4 = 3(cos pi + i sin pi)
  By De Mieve  Law , we have x = 3^(1/4)(cos pi/4 + i sin pi/4)
                               = 3^(1/4)(sqrt(2)/2 + i sqrt(2)/2)
  
  This is one of the 4th (primitive) root of -3.

  Better and complete general solutions as:
  x^4 = 3(cos (2k pi +pi) + i sin (2k pi +pi)) for integer k 
 So, x = 3^(1/4)(cos (2k+1)pi/4 + i sin(2k+1)pi/4) where k =0,1,2,3
  Hence, x = 3^(1/4)(sqrt(2)/2 + i sqrt(2)/2) (when k = 0)
   or x = 3^(1/4)(cos 3pi/4 + i sin 3pi/4)
       = 3^(1/4)(-sqrt(2)/2 + i sqrt(2)/2) (when k = 1)
   or x = 3^(1/4)(cos 5pi/4 + i sin 5pi/4)
       = 3^(1/4)(-sqrt(2)/2 - i sqrt(2)/2) (when k = 2)
   or x = 3^(1/4)(cos 7pi/4 + i sin 7pi/4)
       = 3^(1/4)(sqrt(2)/2 - i sqrt(2)/2) (when k = 3)
 
 For any integer n and complex number w=a+bi, the equation x^n = w 
 has n solutions. 
 By  x^n = w = r(cos t + i sin t) where r = }w} and tan t = b/a
   x = r^(1/n) cos ( 2pik + t)/n + i sin ( 2pik + t)/n , k=0,1,2,..,k-1

 Kenny