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If z is one of the fifth roots of unity, not 1, show that:

$1+z+z^2+z^3+z^4=0$

Which wasn't too bad, but the next part is killing me: show that:

$z-z^2+z^3-z^4=2i(sin(2\pi/5)-sin(\pi/5))$

Can anyone help? Thanks!

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I don't think that the second equation is true for all the fifth roots of unity. If you substitute $z^2$ (if $z$ is a fifth root of unity, then so is $z^2$) that expression gets multiplied by -1. –  Jyrki Lahtonen Mar 18 '12 at 6:51
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As a hint: Have you drawn a picture? Take a look at where these numbers are on the complex plane! If $z=\cos(2\pi/5)+i\sin(2\pi/5)$, then what is $z-z^4$? Similarly $z^2-z^3$? –  Jyrki Lahtonen Mar 18 '12 at 6:57
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2 Answers 2

up vote 4 down vote accepted

For the first part you should note that by the formula for the geometric series,

$$1 + z + z^2 + z^3 + z^4 = \frac{z^5 - 1}{z- 1}.$$

So if you put $z = e^{2\pi i/5}$ into $z^5 - 1$ you get $(e^{2\pi i/5})^5 =1$ so that $1 + z + \ldots z^4 = 0$. You can see also that if you put $z = (e^{2\pi i/5})^n$ for $n=2,3,4$ you see that $z^5 - 1$ is always zero.

For the second part you should be able to see that if $z = e^{2\pi i/5}$, then $$z - z^4 = 2i\sin(2\pi/5).$$

The reason is if you draw a pentagon in the complex plane with vertices at the points $$1, e^{2\pi i/5}, e^{4\pi i/5}, e^{6\pi i/5}, e^{8\pi i/5}$$

then the vertices $e^{2\pi i /5}$ and $e^{8 \pi i/5}$ have equal real part so subtracting one from the other gives the result above. You should be familiar with de Moivre's formula for this.

Now similarly if you calculate $z - z^4$ you should get

$$z^3 - z^2 = 2i\sin(6 \pi/5).$$

But then $2i\sin(6 \pi/5) = -2i\sin(\pi/5)$ so adding this together with that found for $z - z^4$ gives

$$z +z^3 - z^2 -z^4 = 2i(\sin(2 \pi/5) - \sin(\pi/5)).$$

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@jason: if you don't know what $e$ is, how do you write the 5th roots of unity? –  Martin Argerami Mar 18 '12 at 7:51
    
@Jason Do you not know of the formula $e^{i\theta} = \cos \theta + i \sin \theta$??? –  user38268 Mar 18 '12 at 7:53
    
Ahh.. I see. Thanks, Benjamin! Sorry, never seen that notation... it's the same as $cis(\theta)$ right? –  Jason Mar 18 '12 at 7:56
    
@Jason Yes $\operatorname{cis} \theta$ is short hand for $\cos \theta + i \sin \theta$. By the way in $\LaTeX$ if you want to type $cis$ but don't want it in italics you do \operatorname{cis}. –  user38268 Mar 18 '12 at 8:04
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This way is certainly less direct than the previous solution, but more straightforward.

\begin{eqnarray} z-z^2+z^3-z^4&=&-\sum_{k=1}^4(-1)^ke^{2ik\pi/5}=-\sum_{k=1}^4\left(-e^{2i\pi/5}\right)^k=-\frac{-e^{2i\pi/5}-(-e^{2i\pi/5})^5}{1-(-e^{2i\pi/5})}\\ &=&\frac{e^{2i\pi/5}-1}{e^{2i\pi/5}+1}=\frac{(e^{2i\pi/5}-1)(e^{-2i\pi/5}+1)}{(1+\cos2\pi/5)^2+\sin^22\pi/5}=\frac{e^{2i\pi/5}-e^{-2i\pi/5}}{2+2\cos2\pi/5}\\ &=&\frac{i\sin2\pi/5}{1+\cos2\pi/5}=\frac{2i\sin\pi/5\,\cos\pi/5}{2\cos^22\pi/5}= \frac{i\sin\pi/5}{\cos\pi/5}=i\tan\pi/5. \end{eqnarray}

So it remains to check that $\tan\pi/5=2(\sin2\pi/5-\sin\pi/5)$. After a simplification, this amounts to check that $$ 1=2\cos\pi/5\;(2\cos\pi/5-1), $$ and this can be seen directly from the fact that $\cos\pi/5=(1+\sqrt5)/4$.

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