A Complex problem

Algebra Level 3

Let { ϕ i } 1 n − 1 \{\phi_i\}_1^{n-1} denote the set of solutions to the equation : ∑ k = 0 n − 1 z k = 0 , w h e r e n > 1 a n d z ∈ C \sum_{k=0}^{n-1} \ z^k = 0, \quad where \ n >1 \ and \ \ z \in \mathbb C

What is the value of ∑ k = 1 n − 1 ϕ k \displaystyle \sum_{k=1}^{n-1} \ \phi_k ?

n 2 if n is even, and n − 1 2 if n is odd. \frac n2 \text{ if } n \text{ is even, and } \frac{n-1}2 \text{ if } n \text{ is odd.} 1 if n is even, and − 1 if n is odd. 1 \text{ if } n \text{ is even, and } -1 \text{ if } n \text{ is odd.} − 1 -1 − 1 if n is even, and 1 if n is odd. -1 \text{ if } n \text{ is even, and } 1 \text{ if } n \text{ is odd.} 1 1 0 0

Patrick Bourg by Patrick Bourg , 28, United Kingdom

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1 solution

Direct Vieta.....
Sum of the all roots of the given polynomial is -1 \fbox{-1}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

can you translate the bracket portion of the question into english for me? thanks!!!

Willia Chang - 5 years, 1 month ago

Sum of all the roots to the given polynomial equation is 0 0 . However, it can be easily seen that roots to this equation are the first 23 23 imaginary roots of unity from it follows that their sum is − 1 -1 .

Aditya Sky - 4 years, 6 months ago

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