Question

Let ω be a complex number such that ω^7 = 1 and ω ≠ 1. Compute:

ω16+ω18+ω20+⋯+ω54.
A) 1
B) 7
C) 0
D) -7

Answer 1

The sum of the series ω16 to ω54 is 0 because ω is a 7th root of unity, and ω raised to multiples of 7 equals 1, which makes the repeating pattern of ω in the series cancel out. Correct option is C) 0

Step-by-step explanation:

First, let's note that since ω is a complex solution to ω7 = 1, and ω ≠ 1, ω has to be one of the 6 non-real 7th roots of unity. Each of these roots of unity, when raised to the 7th power, gives 1. Therefore, ω7k will always be 1 for any integer k. Using this property, ω16 can be rewritten as ω(7\times2)+2 = ω2 and similarly for the other powers in the series:

• ω16 = ω2
• ω18 = ω4
• ω20 = ω6
• ...
• ω54 = ω5

When we add these together, every term ωn where n is a multiple of 7 will add to 1, and we will be left with a sum of powers of ω that are not multiples of 7. Since ω7 = 1, the pattern of powers of ω will repeat every 7 terms. Since the powers from 16 to 54 cover 7 full cycles (14 to 49) and part of an eighth cycle (50 to 54), the sum of the series will be 0, because the sum within each cycle is 0. The answer to the problem is C) 0.