Question 1071957
Find all the complex solutions of the equation.
x^4 + i = 0
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{{{ x^4 = -i }}}
Use the fact that  -i = e^(3(pi)i/2) :
 x^4 = e^(3(pi)i/2) 
Here's where its a little tricky - to find all the solutions, we must account for the fact that 
e^w = e^(w+2(pi)k) for k=0,1,2,…  (i.e. the exponential aliases on top of itself every 2(pi))
 
Re-writing:
 x^4 = e^((3(pi)/2) + 2k(pi))i 
Raise both sides to the 1/4 power:
 (x^4)^(1/4) = e^(((3(pi)/2) + 2k(pi))i * (1/4))
    x = e^((3(pi)/8 + k*(pi)/2)i)
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Using Euler's equation:  e^(ni) = cos (n) + i*sin(n),  we re-write the above:
--
  x = cos((3(pi)/8 + k(pi)/2) + i*sin(3(pi)/8 + k(pi)/2)

k=0:   x = cos(3(pi)/8) + i*sin(3(pi)/8)    
 [ approx:  x = 0.38268343 + i*0.92387953 ]
—
k=1:   x = cos(3(pi)/8 + (pi)/2) + i*sin(3(pi)/8 + (pi)/2)  
 [ approx: x = -0.92387953 + i*0.38268343 ]
—
k=2:   x = cos(3(pi)/8 + (pi)) + i*sin(3(pi)/8 + (pi))   
   [ approx:  x = -0.38268343 - i*0.92387953 ] 
—
k=3:   x = cos(3(pi)/8 + 3(pi)/2) + i*sin(3(pi)/8 + 2(pi)/2)      
  [ approx:  x = 0.92387953 - i*0.38268343  ] 
—
When k reaches 4, the 2nd term in cos( ) and sin( ) reaches 2(pi) which means we've wrapped around once so the above four answers are the unique solutions (the rest are aliases).