Final answer:
The fourth roots of -3 + 3√3i are 1 + i, -1 - i, -1 + i, and 1 - i.
Step-by-step explanation:
To find the fourth roots of -3 + 3√3i, we'll first express the complex number in polar form. Let z = -3 + 3√3i. The modulus of z, |z|, can be calculated as the square root of the sum of squares of the real and imaginary parts: |z| =
= √(9 + 27) = √36 = 6. The argument of z, θ, can be found using the inverse tangent function: θ =
= -π/3.
Now that we have z in polar form (z = 6 * (cos(-π/3) + i * sin(-π/3))), the fourth roots can be obtained by taking the nth root of the modulus and dividing the argument by n. For fourth roots, n = 4. So, the fourth root of z is given by
* [cos((-π/3 + 2kπ) / 4) + i * sin((-π/3 + 2kπ) / 4)], where k = 0, 1, 2, 3.
Substituting k = 0, 1, 2, 3 into the formula, we get:
1st root:
* [cos((-π/3) / 4) + i * sin((-π/3) / 4)] = 1 + i
2nd root:
* [cos((5π/12) + i * sin(5π/12)] = -1 - i
3rd root:
* [cos((7π/12) + i * sin(7π/12)] = -1 + i
4th root:
* [cos((11π/12) + i * sin(11π/12)] = 1 - i
Therefore, the fourth roots of -3 + 3√3i are 1 + i, -1 - i, -1 + i, and 1 - i.