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The complex cube roots of 13+13i?

The complex cube roots of 13+13i?-example-1
User Ncrocfer
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ω(13 + 13i) = 2 + √13 i

ω^2(13 + 13i) = -1 + √13 i

ω^3(13 + 13i) = 13 - 13i

ω(13 + 13i) = 2 + √13 i, where ω is a complex cube root of 1.

ω^2(13 + 13i) = -1 + √13 i.

ω^3(13 + 13i) = 13 - 13i.

These can be found using DeMoivre's Theorem, which states that for any complex number z and any positive integer n:

(cis θ)^n = cis nθ

where cis θ is the complex number in polar form with angle θ.

In this case, z = 13 + 13i and n = 3. Therefore, the cube roots of z are:

ωz = cis(θ/3), ω^2z = cis(2θ/3), and ω^3z = cisθ.

The angle θ can be found by solving the equation z = cis θ. In this case, z = 13 + 13i has angle θ = arctangent(13/13) = π/4.

Therefore, the cube roots of z are:

ωz = cis(π/12), ω^2z = cis(π/2), and ω^3z = cis(π/4).

These can then be converted to rectangular form using the following formulas:

cis θ = cos θ + i sin θ

sin θ = 2 / (1 + e^(-2iθ))

cos θ = (1 + e^(-2iθ)) / 2

User QCring
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