Complex Numbers
- Thread starter ashive5i
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Cubist
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x^4 = 1
x^4 - 1 = 0
Factor to obtain...
(x^2 + 1)*(x^2 - 1) = 0
(x^2 + 1)*(x + 1)*(x - 1) = 0
Therefore there are multiple solutions to x^4 = 1. Specifically x={1,-1,i} where i=√-1
Generally when you square both sides of an equation you introduce more solutions. So after squaring it often makes sense to go back to the original equation to see which of the solutions fit the original problem, and eliminate any solutions that were introduced due to squaring.
EDIT: I forgot about the "-i" solution pointed out by HallsofIvy below
x^4 - 1 = 0
Factor to obtain...
(x^2 + 1)*(x^2 - 1) = 0
(x^2 + 1)*(x + 1)*(x - 1) = 0
Therefore there are multiple solutions to x^4 = 1. Specifically x={1,-1,i} where i=√-1
Generally when you square both sides of an equation you introduce more solutions. So after squaring it often makes sense to go back to the original equation to see which of the solutions fit the original problem, and eliminate any solutions that were introduced due to squaring.
EDIT: I forgot about the "-i" solution pointed out by HallsofIvy below
HallsofIvy
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First, since there are two numbers, i and -i, in the complex numbers, whose square is -1, it is better to write your first line as "x= i".
Your error is where you go from "\(\displaystyle x^4= 1\)" to "\(\displaystyle x= 1\)". \(\displaystyle x^4= 1\) is a fourth degree polynomial so has four solutions in the complex numbers, 1, -1, i, and -i.
Your error is where you go from "\(\displaystyle x^4= 1\)" to "\(\displaystyle x= 1\)". \(\displaystyle x^4= 1\) is a fourth degree polynomial so has four solutions in the complex numbers, 1, -1, i, and -i.