Given a complex number \(\displaystyle\omega={\cos{{\left({\frac{{{2}\pi}}{{{5}}}}\right)}}}+{i}{\sin{{\left({\frac{{{2}\pi}}{{{5}}}}\right)}}}\)

Adolfo Hebert

Adolfo Hebert

Answered question


Given a complex number ω=cos(2π5)+isin(2π5)

Answer & Explanation

Wilson Rivas

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 nN, and consider the complex nth root of 1, i.e., the complex number
that satisfies the equation
Of course there are n solutions to this equation, we take ωn as the one with the minimal argument (except from 1). The calculation of ωnn is as you have done - using de Moivre formula,
By the sum of a geometric sequence (which is also true for complex numbers), we know that
and therefore plugging in z=ωn yields zero in the RHS, and the sum you were requested to calculate in the LHS.

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