SOLUTION: Find all the complex solutions of the equation. x^4 + i = 0

Question 1071957: Find all the complex solutions of the equation.
x^4 + i = 0
Answer by math_helper(2461) (Show Source): You can put this solution on YOUR website!
Find all the complex solutions of the equation.
x^4 + i = 0
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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)
--
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 ]
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k=1: x = cos(3(pi)/8 + (pi)/2) + i*sin(3(pi)/8 + (pi)/2)
[ approx: x = -0.92387953 + i*0.38268343 ]
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k=2: x = cos(3(pi)/8 + (pi)) + i*sin(3(pi)/8 + (pi))
[ approx: x = -0.38268343 - i*0.92387953 ]
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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 ]
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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).

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