Question 322660: I understand how to find the square root of i, but I am unsure how to find the cube root of i
the following problem is an example
show that sqrt(3)/2+i/2 is a cube root of i
thank you!
Found 2 solutions by stanbon, CharlesG2:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! show that sqrt(3)/2+i/2 is a cube root of i
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[[sqrt(3)+i]/2]^3
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[(sqrt(3)+i)/2]^2 /2
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= [[3-1+2sqrt(3)i]/4]/2
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= [2+2sqrt(3)i]/4 /2
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= (1+sqrt(3)i)/2]/2
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= [sqrt(3) + i + 3i -sqrt(3)]/4
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= 4i/4
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= i
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cheers,
Stan H.
Answer by CharlesG2(834) (Show Source):
You can put this solution on YOUR website! "I understand how to find the square root of i, but I am unsure how to find the cube root of i
the following problem is an example
show that sqrt(3)/2+i/2 is a cube root of i
thank you!"
i is defined as the square root of -1, i is an imaginary number
i^2 = -1
a complex number is a number consisting of a real part and an imaginary part,
it is written in the form a+bi
i has 2 roots +i and -i, -i is the same as -1 * i, (+i)^2 = i^2 = -1,
(-i)^2 = i^2 = -1
now we want to find the cube root of i,
this would be the same as x^3 = i, where x is the cube root that we are solving for
let x = a + bi
(a + bi)^3 = i
(a + bi)(a + bi)(a + bi) = i (use FOIL)
(a + bi)(a^2 + 2abi + b^2 * i^2) = i
(a + bi)(a^2 + 2abi - b^2) = i (distribute)
a^3 + 2bia^2 - ab^2 + bia^2 + 2ab^2 * i^2 - b^3 * i = i
a^3 + 2bia^2 - ab^2 + bia^2 - 2ab^2 - b^3 * i = i
a^3 - ab^2 - 2ab^2 + 2bia^2 + bia^2 - b^3 * i = i
a^3 - 3ab^2 + 3 * a^2 * b * i - b^3 * i = i
(a^3 - 3ab^2) + (3a^2 * b - b^3)i = i
show that sqrt(3)/2 + i/2 is a cube root of i
a = sqrt(3)/2, b = 1/2
the following part needs to equal 0:
a^3 - 3ab^2 = 0 (plug in values for a and b)
(sqrt(3)/2)^3 - 3(sqrt(3)/2)(1/2)^2 = 0
(3sqrt(3))/8 - (3sqrt(3))/2 * 1/4 = 0
(3sqrt(3))/8 - (3sqrt(3))/8 = 0
0 = 0
the following part needs to equal 1:
(3a^2 * b - b^3) = 1 (plug in values for a and b)
3 * (sqrt(3)/2)^2 * 1/2 - (1/2)^3 = 1
3 * 3/4 * 1/2 - 1/8 = 1
9/8 - 1/8 = 1
8/8 = 1
1 = 1
so:
sqrt(3)/2 + i/2 is indeed a cube root of i
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