Question

If ω is a complex cube root of unit, then 1 + ω + ω² + ... + ω¹⁰⁰ is equal to

Answer 1

Correct Answer - Option 2 : 1 + ω

Concept:

Sum of GP series of nth terms:

\(S_n=\frac{a{*({r}^{n}-1)}}{r-1}\)

Where,

a: First Term

r: common ratio

If ω is a complex root of the unit. Then,

ω³ = 1

Explanation:

According to question 

1 + ω+ ω2 + ... + ω100'

We know that,

ω3 = 1

so, (ω3)33= (1)33

ω99 = 1 --- (1)

Now, Solving GP:

\(= \frac{1{\times({ω}^{101}-1)}}{ω-1}\)

From equation (1);

\(=\frac{\omega^{99}(\omega^2-1)}{\omega-1}\)

\(=\frac{\omega^{99}(\omega-1)(\omega+1)}{\omega-1}\)

= ω + 1