We have the complex number:
z = 27(cos 279° + i sin 279°)
The modulus of the complex number is r = 27, and the argument (angle) is θ = 279°.
To find the cube roots, we use the formula for the nth roots of a complex number:
w_k = r^{1/n} (cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n))
where n = 3 (since we are finding cube roots), k = 0, 1, 2.
First, we calculate the modulus of the cube roots:
r^{1/3} = 27^{1/3} = 3
Next, we calculate the angles for each cube root:
For k = 0:
- The angle is: (279° + 2(0)π)/3 = 279°/3 = 93°
For k = 1:
- The angle is: (279° + 2π)/3 = (279° + 360°)/3 = 639°/3 = 213°
For k = 2:
- The angle is: (279° + 4π)/3 = (279° + 720°)/3 = 999°/3 = 333°
Now, we can express the cube roots:
- For k = 0: w_0 = 3(cos 93° + i sin 93°)
- For k = 1: w_1 = 3(cos 213° + i sin 213°)
- For k = 2: w_2 = 3(cos 333° + i sin 333°)
Thus, the three cube roots of the given complex number are:
- w_0 = 3(cos 93° + i sin 93°)
- w_1 = 3(cos 213° + i sin 213°)
- w_2 = 3(cos 333° + i sin 333°)
These roots are equally spaced around the unit circle in the complex plane, each separated by 120°.