Question

How do I find the 5 roots of 1^(1/5) using De Moivre's Theorem?​

Answer 1

Do you mean the 5th roots of 1? Or the 5th roots of 1^(1/5), i.e. the 5th roots of the 5th roots of 1?

I assume you mean the former. To begin with,

1 = exp(0°i ) = cos(0°) + i sin(0°)

Take the 5th root, i.e. (1/5)th power of both sides, so that by DeMoivre's theorem,

1^(1/5) = [ cos(0°) + i sin(0°) ]^(1/5)

1^(1/5) = cos((0° + 360°n)/5) + i sin((0° + 360°n)/5)

where n = 0, 1, 2, 3, or 4. Then 1^(1/5) has the 5 possible values,

• cos(0°/5) + i sin(0°/5) = 1

• cos(360°/5) + i sin(360°/5) = cos(72°) + i sin(72°)

• cos(720°/5) + i sin(720°/5) = cos(144°) + i sin(144°)

• cos(1080°/5) + i sin(1080°/5) = cos(216°) + i sin(216°)

• cos(1440°/5) + i sin(1440°/5) = cos(288°) + i sin(288°)

If you wish, or are required to, you can go on to write these in terms of radicals by expanding cos(5x) and sin(5x) (where x = 72° so that 5x = 360°). For example,

cos(5x) = cos⁵(x) - 10 cos³(x) sin²(x) + 5 cos(x) sin⁴(x)

cos(5x) = cos⁵(x) - 10 cos³(x) (1 - cos²(x)) + 5 cos(x) (1 - cos²(x))²

cos(5x) = 16 cos⁵(x) - 20 cos³(x) + 5 cos(x)

which follows from the expansion

(cos(x) + i sin(x))⁵ = cos(5x) + i sin(5x)

due to DeMoivre's theorem and equating the real parts. Then cos(5x) = 1, and you can try to solve for cos(x) :

1 = 16 cos⁵(x) - 20 cos³(x) + 5 cos(x)

Actually, it would be easier to find sin(x) first, since sin(5x) = 0. The expansion in terms of sin(x) will look quite similar to the one shown here.