Nth root of a complex number should return N results

General Usage
1

As shown below:

x = 27 * e^π
e^2im
e^(2+2pi)im
n = 3
Θ = pi/3
Q = 27
x = Q⋅e^(Θ⋅im)
ϕ₁ = Θ/n + 2π⋅1/n
ϕ₂ = Θ/n + 2π⋅2/n
ϕ₃ = Θ/n + 2π⋅3/n
R = Q^(1/n)
y₁ = R⋅e^(ϕ₁⋅im)
y₂ = R⋅e^(ϕ₂⋅im)
y₃ = R⋅e^(ϕ₃⋅im)
println(x)
println(y₁^3)
println(y₂^3)
println(y₃^3)
println(x^(1/3))
println(y₁)
println(y₂)
println(y₃)

I was expecting x^(1/3) to give me 3 roots, but it only returned one of them. Why? How can I get the others?

2

x ^ n is the exponentiation operation, not the root finder. If you want to find all the roots of a complex number, you can for example search for the roots of the corresponding polynomial using PolynomialRoots.jl (but there are also other packages for looking for roots of polynomials, like Polynomials.jl):

julia> using PolynomialRoots

julia> roots([8, 0, 0, -1])
3-element Array{Complex{Float64},1}:
         -1.0-1.73205im
 2.0-9.61481e-17im     
         -1.0+1.73205im

The argument of roots is the vector of coefficients of the polynomial, from the lowest to the highest one. This gives you the 3rd roots of 8.

In your case:

julia> roots([x, 0, 0, -1])
3-element Array{Complex{Float64},1}:
   2.81908+1.02606im
 -0.520945-2.95442im
  -2.29813+1.92836im

Note also that for the exponentiation operation x ^ (1/3) with x real you can use instead cbrt(x), which is often faster and more accurate.

3 Likes
3

PolynomialRoots.jl is very nice!

4

PolynomialRoots.jl is indeed very nice, but possibly overkill for this problem, as something like

function allroots{T <: Real}(x::Complex{T}, n::Int)
    z = x^(1/n)
    [z*cis((i-1)*2*π/n) for i in 1:n]
end

should get all the roots.

5 Likes