cube root

Kiran · 2006-12-19 11:19:17

find the three cube roots of 1 and express tham in rectangular coordinates.

luca-deltodesco · 2006-12-19 19:27:56

according to wikipedia, that might help you to find them so you can see whether youre on the right track, i wouldnt know how to find them myself.

mathsyperson · 2006-12-20 00:35:12

It's probably easiest to find the cube roots in modulus-argument form first, then convert them to rectangular co-ordinates later.

De Moivre's Theorem says that [a(cos θ + isin θ)]^n = a^n(cos nθ + isin nθ).

This means that 1^(1/3) = (1+0i)^(1/3) = [1(cos 0 + isin 0)]^(1/3).

But because trigonometry cycles, we can also say that 1^(1/3) = (cos 2π + isin 2π)^(1/3) and that 1^(1/3) = (cos 4π + isin 4π)^(1/3).

Therefore, using De Moivre's Theorem, the cube root of 1 is given by (cos 0 + isin 0), (cos 2π/3 + isin 2π/3) and (cos 4π/3 + isin 4π/3).

In rectangular coordinates, these are 1, -1/2 + √(3/2) and -1/2 - √(3/2), respectively.

If you plot these points on the Complex Plane, they form an equilateral triangle on the unit circle, with one point on the real line.

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