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Question 1066706: Solve the equation (z-1)^3 + ((i-1)/(i+1))^3 =0
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! (z - 1)^3 + ((i - 1) / (i + 1))^3 = 0
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Note that ((i - 1) / (i + 1))^3 = -i, since (i - 1) / (i + 1) = i then i^3 = i^2 * i = -1 * i = -i
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(z - 1)^3 -i = 0
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(z - 1)^3 = i
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Note i = (-i)^3 and let x = (z - 1), then
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x^3 - (-1)^3 = 0
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we know that a^3 - b^3 = (a - b) * (a^2 +ab + b^2), then
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{x−(−i)} * {x^2+x(−i)+(−i)^2} = {x−(−i)} * (x^2−ix−1) = 0
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If x−(−i) = 0, x = −i, now x = z - 1 then
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z - 1 = -i
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z = 1 - i
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Else x^2 −ix −1 = 0
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use the quadratic formula
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x = (i + or - square root(i^2 -(4)(-1))) / 2 = (i + or - square root(3)) / 2)
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z - 1 = (i + or - square root(3)) / 2)
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z = 1 + (i + or - square root(3)) / 2
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Note there are 3 solutions for z
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z = 1 - i
z = 1 + (i + square root(3)) / 2
z = 1 + (i - square root(3)) / 2
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