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Let $w = e^{i\frac{2\pi}5}$. I would like to evaluate

$$w^0 + w^1 + w^2 + w^3 +...+ w^{49}$$

Can anyone please give me an idea how to evaluate the expression?

Thanks in advance

user78723
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4 Answers4

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Recall geometric series: $$1+w+w^2 + \cdots + w^n = \dfrac{1-w^{n+1}}{1-w}$$ Also, make use of the fact that $$e^{i(2k\pi + \theta)} = e^{i \theta},\,\,\,\,\, \text{where } k \in \mathbb{Z}$$ I trust you can finish it off from here.

  • Thanks. I found the same equation than you, but I don't know how to replace the relation e^{i(2k\pi + \theta)} = e^{i \theta},,,,,, \text{where } k \in \mathbb{Z} in the sum equation. – user78723 May 21 '13 at 05:12
  • @user78723 We have $w^{49+1} = w^{50} = e^{2 \pi i/5 \times 50} = e^{20 \pi i} = 1$. Hence, the numerator is $1-w^{50} = 1 -1 = 0$. –  May 21 '13 at 05:58
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Note that $e^{2\pi i k/5}$, where $k$ ranges from $0$ to $4$, gves us all the roots of the equation $z^5-1=0$. The sum of these $5$ roots is the coefficient of $z^4$, which is $0$.

André Nicolas
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  • I calculated the first 10 terms and I arrived to zero, but I don't know to show this for the entire sum. – user78723 May 21 '13 at 05:05
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    You only need the first five, which we know. The next $5$ is $w^5+w^6+\cdots +w^9=w^5(w^0+w^1+\cdots+w^4)=(w^5)(0)=0$. The next $5$ after that are $w^{10}(w^0+\cdots +w^4)=0$. The next bunch are $w^{15}(w^0+\cdots +w^4)=0$. And so on. – André Nicolas May 21 '13 at 05:10
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This is the sum of the fifth roots of unity, ten times over. See here for a discussion of why each of these sums is zero.

Community
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vadim123
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Hints:

$A=w^0 + w^1 + w^2 + w^3 +...+ w^{49}(w-1)=w^{50}-1=(w^5)^{10}-1=(w^5-1)( \dots$

$w=e^{i \frac{2 \pi k}{5}} \implies1+ w+w^2+w^3+w^4=0 \implies A=0$

Inceptio
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