Explained: "Given a complex number ω=cos⁡(2π5)+isin⁡(2π5)"

Adolfo Hebert

Answered question

2022-03-25

Given a complex number

ω = \cos (\frac{2 π}{5}) + i \sin (\frac{2 π}{5})

Wilson Rivas

Beginner2022-03-26Added 12 answers

Step 1
You seem to have written a perfectly valid proof. Just note that the general case is also true (and pretty important as well). Let

n \in N

, and consider the complex nth root of 1, i.e., the complex number

ω_{n} = e^{\frac{2 π t}{n}} = \cos (\frac{2 π}{n}) + i \sin (\frac{2 π}{n})

that satisfies the equation

z^{n} = 1

Of course there are n solutions to this equation, we take

ω_{n}

as the one with the minimal argument (except from 1). The calculation of

ω_{n}^{n}

is as you have done - using de Moivre formula,

ω_{n}^{n} = {(e^{\frac{2 π t}{n}})}^{n} = e^{2 π} = 1

By the sum of a geometric sequence (which is also true for complex numbers), we know that

1 + z + z^{2} + \dots + z^{n - 1} = \frac{z^{n} - 1}{z - 1}

and therefore plugging in

z = ω_{n}

yields zero in the RHS, and the sum you were requested to calculate in the LHS.

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