To find the complex cube roots of 8, we first express 8 in polar form. The number 8 can be written as:
8 = 8 + 0i
In polar form, this can be represented as:
8 = 8 (cos 0 + i sin 0)
Here, the modulus (r) is 8, and the argument (θ) is 0. We want to find the complex cube roots, which means we are looking for solutions to:
w³ = 8
To find the cube roots, we use the formula:
w_k = r^{1/n} (cos(θ/n + 2kπ/n) + i sin(θ/n + 2kπ/n))
In our case, n = 3 (since we want cube roots) and k can be 0, 1, or 2.
Calculating the modulus:
r^{1/3} = 8^{1/3} = 2
Now, we calculate the argument for each value of k:
For k = 0:
θ/3 + 2π(0)/3 = 0/3 + 0 = 0
w_0 = 2 (cos 0 + i sin 0) = 2
For k = 1:
θ/3 + 2π(1)/3 = 0/3 + 2π/3 = 2π/3
w_1 = 2 (cos(2π/3) + i sin(2π/3)) = 2 (-1/2 + i √3/2) = -1 + i√3
For k = 2:
θ/3 + 2π(2)/3 = 0/3 + 4π/3 = 4π/3
w_2 = 2 (cos(4π/3) + i sin(4π/3)) = 2 (-1/2 – i √3/2) = -1 – i√3
So, the three complex cube roots of 8 are:
- w_0 = 2
- w_1 = -1 + i√3
- w_2 = -1 – i√3