Let z1=2(cos 2pi/3+isin 2pi/3)

Are you familiar with DeMoivre's Theorem?

no not at all

Too bad,

ok, the long way then .....

I am just going to call z1 as z

let z^(1/3) = a + bi
then z = (a+bi)^3
= a^3 + 3a^2 bi + 3a b^2 i^2 + b^3 i^3
= a^3 + 3a^2 b i - 3ab^2 -b^3 i
= (a^3 - 3ab^2) + i (3a^2b - b^3)

but z = 2(cos 2π/3 + i sin 2π/3) = -1 + √3i

so by comparison
a^3 - 3ab^2 = -1 , and
3a^2 - b^3 = √3

nasty , nasty to solve, so I ran it through Wolfram
using x and y instead of a and b
http://www.wolframalpha.com/input/?i=+x%5E3+-+3xy%5E2+%3D+-1+%2C+3x%5E2+-+y%5E3+%3D+%E2%88%9A3
notice we have more than one solution, one such solution would be
.86363 + .796609 i

what are those numbers in radians

To calculate the value of z1^(1/3), we need to express the complex number z1 in its polar form.

Given z1 = 2(cos(2π/3) + isin(2π/3)), we can rewrite it in polar form as:
z1 = 2e^(i(2π/3))

To find the cube root of z1, we need to find the cube root of its magnitude and divide the argument by 3.

Magnitude of z1:
|z1| = √(cos^2(2π/3) + sin^2(2π/3))
= √(1)
= 1

Argument of z1:
Arg(z1) = arctan(sin(2π/3) / cos(2π/3))
= π/3

Now, let's calculate the cube root of z1:
Cube root of the magnitude:
∛|z1| = ∛(1) = 1

Cube root of the argument:
∛Arg(z1) = (∏/3) / 3 = ∏/9

So, z1^(1/3) = 1*cos(∏/9) + i*sin(∏/9)