# Sum of the e's

## Homework Statement

hi,
I'm working on constructible things again and in one of the proofs our prof threw out this identity and I just don't know where it came from:
$$1+e^{\frac{2\pi}{7}i}+e^{\frac{4\pi}{7}i}+e^{\frac{6\pi}{7}i}+e^{\frac{8\pi}{7}i}+e^{\frac{10\pi}{7}i}+e^{\frac{12\pi}{7}i}=\frac{e^(\frac{12\pi}{7}i)^7-1}{e^{\frac{12\pi}{7}i}-1}$$
HOW did he get that?

Edit: I can't tell if the final term is supposed to be 2pi of 12 pi. I dunno.

Last edited:

## Answers and Replies

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Curious3141
Homework Helper
Hint : geometric progression.

The exponents on the Right Hand Side should be 2pi, not 12 pi.

$$x^n-1=(x-1)(1+x+x^2+x^3+x^4+x^5+...x^{n-1})$$

for n=natural number.

Thanks. I see it now.
CC