find the fourth roots of 1-root3i

To find the fourth roots of 1 - root 3i, we need to find the values of x such that x^4 = 1 - root 3i.

Step 1: We need to convert 1 - root 3i into its trigonometric form.
Let's assume x = a + bi, where a and b are real numbers.

So, x^4 = (a + bi)^4 = a^4 + 4a^3bi + 6a^2b^2 - 4ab^3i + b^4i^4
Since i^2 = -1 and i^4 = 1, we can simplify this to:
x^4 = (a^4 - 6a^2b^2 + b^4) + (4a^3b - 4ab^3)i

Now, equating the real and imaginary parts, we get:
a^4 - 6a^2b^2 + b^4 = 1
4a^3b - 4ab^3 = -root 3

Step 2: Solve the equations for a and b.
From the first equation, we can rewrite it as:
(a^2 - b^2)^2 - 4a^2b^2 = 1
Let's substitute u = a^2 - b^2:
u^2 - 4a^2b^2 = 1

Simplifying the second equation, we get:
a(a^2 - b^2) = -root 3
Substitute u = a^2 - b^2:
(2u + b^2)(u) = -root 3

Now, we have two equations in terms of u and b.
Solving these equations can be quite complicated.

Step 3: Alternative method using De Moivre's Theorem:
We can use De Moivre's theorem to directly find the fourth roots of 1 - root 3i.

De Moivre's theorem states:
(r * (cos θ + i sin θ))^n = r^n * (cos (nθ) + i sin (nθ))

Let's apply this to our case:
1 - root 3i = 2 * (cos (π/3) - i sin (π/3))

When we have a number in the form r * (cos θ + i sin θ), its nth root is given by:
r^(1/n) * (cos (θ/n + 2πk/n) + i sin (θ/n + 2πk/n)), where k is an integer.

For our case, we have:
r = 2, θ = π/3, n = 4

The fourth roots of 1 - root 3i are:
(2^(1/4)) * (cos (π/12 + 2πk/4) + i sin (π/12 + 2πk/4)), where k is an integer.

Simplifying further, we have:
(2^(1/4)) * (cos (π/12 + πk/2) + i sin (π/12 + πk/2)), where k is an integer.

Therefore, the fourth roots of 1 - root 3i are:
(2^(1/4)) * (cos (π/12), sin (π/12)), (2^(1/4)) * (cos (5π/12), sin (5π/12)), (2^(1/4)) * (cos (9π/12), sin (9π/12)), (2^(1/4)) * (cos (13π/12), sin (13π/12)).