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Can anyone help me with this question?

"Use De Moivre's theorem to solve the equation $z^5 = 1.$ Show that the points representing the five roots of this equation on an Argand diagram form the ventics of a regular pentagon of Perimeter $10\sin(\pi/5)$."

I manage to solve the roots as $\operatorname {cis} (2k\pi/5)$ where $k$ is an integer from 0 to 4, but how do we derive the perimeter?

Please help; Thanks in advance.

amWhy
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Justin HT
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  • If you have the roots already, just calculate distances? – GPerez Jan 06 '15 at 13:37
  • thanks for the quick response, is that means I should convert it back to a+bi form and then use the distant formula(Pythagoras theorem)? – Justin HT Jan 06 '15 at 13:40
  • Break the pentagon up into five triangles (all touching the origin) and use the cosine rule to get the length of one side of the pentagon (notice you have the length of two sides and the interior angle of the triangle from the roots) – Kelvin Soh Jan 06 '15 at 13:41
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    @JustinHT That's what I would have done. Though, what Kelvin Soh suggests might be easier. Try both! – GPerez Jan 06 '15 at 13:43
  • @KelvinSoh I tried to that method but I couldn't arrive at 10*sin(pi/5)here's what I did:

    since the moduli are 1 and the angle of the triangle are 2pi/5:

    c = sqrt(1+1-((1/2)*cos(2pi/5)) and I have no clue where to go from there...

     – Justin HT Jan 06 '15 at 13:45
  • It should be $1+1-2\cos \ldots$ instead of half. Then we can apply the double angle formula $\cos \frac{2\pi}{5} = 1 - 2 \sin^2 \frac{\pi}{5}$ (next time you see an angle half of or double of what you have the double angle formulas are especially useful) – Kelvin Soh Jan 06 '15 at 13:47
  • oh it must be that mistake (2 instead of 1/2)! Thank you so much for your help! @KelvinSoh – Justin HT Jan 06 '15 at 13:58
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    @Kelvin Soh: Simpler: the basis of each of these triangles measures $2\sin\dfrac\pi5$, from the very definition of sine on the trigonometric circle. – Bernard Jan 06 '15 at 14:00

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