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Let $\omega$ be a complex number such that $\omega^3 = 1$, $\omega \neq 1$, and $S = 1 + \omega + \omega^2 + \ldots + \omega^{n-1}$ - Leaving Cert Mathematics - Question b - 2015

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Let-$\omega$-be-a-complex-number-such-that-$\omega^3-=-1$,-$\omega-\neq-1$,-and-$S-=-1-+-\omega-+-\omega^2-+-\ldots-+-\omega^{n-1}$-Leaving Cert Mathematics-Question b-2015.png

Let $\omega$ be a complex number such that $\omega^3 = 1$, $\omega \neq 1$, and $S = 1 + \omega + \omega^2 + \ldots + \omega^{n-1}$. Use the formula for the sum of a... show full transcript

Worked Solution & Example Answer:Let $\omega$ be a complex number such that $\omega^3 = 1$, $\omega \neq 1$, and $S = 1 + \omega + \omega^2 + \ldots + \omega^{n-1}$ - Leaving Cert Mathematics - Question b - 2015

Step 1

Use the formula for the sum of a finite geometric series

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Answer

The formula for the sum of a finite geometric series is given by:

S=a(1−rn)1−rS = \frac{a(1 - r^n)}{1 - r}

where:

  • aa is the first term of the series
  • rr is the common ratio

In our case, we have:

  • a=1a = 1
  • r=ωr = \omega
  • The number of terms, n=3n = 3 (since ω3=1\omega^3 = 1)

Thus, we can plug these values into the formula:

Step 2

Calculate $S$ using the formula

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Answer

Substituting the values into the formula yields:

S=1(1−ω3)1−ωS = \frac{1(1 - \omega^3)}{1 - \omega}

Since ω3=1\omega^3 = 1, this becomes:

S=1(1−1)1−ω=01−ω=0S = \frac{1(1 - 1)}{1 - \omega} = \frac{0}{1 - \omega} = 0

Therefore, the final value of SS in its simplest form is:

S=0S = 0

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