Prove that the sum of cube roots of unity is zero.

The cube roots of unity are the complex numbers that satisfy the equation:

z^3 = 1

There are three cube roots of unity, which can be written as:

ω1 = 1
ω2 = e^(2πi/3)
ω3 = e^(4πi/3)
Here, i is the imaginary unit (i^2 = -1), and e is the base of the natural logarithm (approximately 2.71828).

Now, let's calculate the sum of these cube roots of unity:

ω1 + ω2 + ω3 = 1 + e^(2πi/3) + e^(4πi/3)

To simplify this sum, we can use Euler's formula, which states:

e^(ix) = cos(x) + i * sin(x)

Using Euler's formula, we can express e^(2πi/3) and e^(4πi/3) as:

e^(2πi/3) = cos(2π/3) + i * sin(2π/3)
e^(4πi/3) = cos(4π/3) + i * sin(4π/3)

Now, let's substitute these values back into the sum:

ω1 + ω2 + ω3 = 1 + (cos(2π/3) + i * sin(2π/3)) + (cos(4π/3) + i * sin(4π/3))

Now, let's separate the real and imaginary parts of the sum:

Real part:
1 + cos(2π/3) + cos(4π/3)

Imaginary part:
i * (sin(2π/3) + sin(4π/3))

Now, let's calculate the real and imaginary parts separately:

Real part:
1 + cos(2π/3) + cos(4π/3)

Using trigonometric identities, we can simplify the real part:

cos(2π/3) = -1/2
cos(4π/3) = -1/2

So, the real part becomes:

1 - 1/2 - 1/2 = 0

Imaginary part:
i * (sin(2π/3) + sin(4π/3))

Using trigonometric identities, we can simplify the imaginary part:

sin(2π/3) = √3/2
sin(4π/3) = -√3/2

So, the imaginary part becomes:

i * (√3/2 - √3/2) = i * 0 = 0

Therefore, the sum of the cube roots of unity is indeed 0. This proves the statement:

ω1 + ω2 + ω3 = 0