Find the sum of the 5th roots of unity?

1 Answer
Mar 7, 2018

\qquad \qquad \qquad \qquad "the sum of the" \ \ 5^{"th"} \ \ "roots of unity is:" \qquad 0.

Explanation:

"By definition, the" \ \ 5^{"th"} \ \ "roots of unity are the solutions of the"
"equation:"

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad x^5 \ = \ 1. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ (I)

"And so they are also the solutions of the equation:"

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad x^5 - 1 \ = \ 0. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (II)

"However, the sum of the roots of the eqn. (II), as with any"
"monic polynomial, is the opposite of the coefficient of the"
"next-to-leading term. In eqn. (II), the next-to-leading term"
"is the" \ \ x^4 \ \ "term. Its coefficient is clearly" \ \ 0. \ \ "So its opposite is"
"clearly" \ \ 0. \ \ "And thus the sum of the solutions of (II) is" \ \ 0. \ \
"Thus the sum of the solutions of (I) is" \ \ 0. \ \ "Thus:"

\qquad \qquad \qquad \qquad "the sum of the" \ \ 5^{"th"} \ \ "roots of unity is:" \qquad 0.

"Some Additional Remarks."

"1) The above argument, thankfully trim, additionally shows:"

\qquad \quad "the sum of the" \ \ n^{"th"} \ \ "roots of unity is:" \qquad 0; \qquad "for" \ \ n >= 2.

\qquad \quad "the sum of the" \ \ 1^{"st"} \ \ "roots of unity is:" \qquad \ 1; \qquad "trivially."

"2) The product of the roots of any monic polynomial of"
"degree" \ \ n, "is" \ \ (-1)^n ( "constant term" ). \ \ "Applying this result to"
"the above polynomial, gives:"

"the product of the" \ \ n^{"th"} \ \ "roots of unity is:" \qquad -1; \qquad "for" \ \ n \quad "odd";

"the product of the" \ \ n^{"th"} \ \ "roots of unity is:" \qquad +1; \qquad "for" \ \ n \quad "even".