The question I am working on starts of with:

Find the five fifth roots of unity and hence solve the following problems

I have done that and solved several questions using this, however when I came to the last question (the one in the title) I got stumped.

The five fifth root of unity are $z= \cos({2\pi k \over 5})+i\sin({2\pi k \over 5}), k\in \mathbb{Z}$ or more simply put $w=\cos({2\pi \over 5})+i\sin({2\pi \over 5})$ and the roots are $z=1, w, w^2, w^3, w^4$.

Now, using this information we are supposed to find the roots of $(z+1)^5=(z-1)^5$. However having tried some different approaches I don't know how to proceed. I tried simplifying it to $5z^4+10z^2+1=0$ from which I suppose I can use the quadratic equation, but it does not utilize the five fifth roots of unity and so I wont get my answer in terms of $w$ (which is the requirement).

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    $\begingroup$ Try rewriting as $ \left(\frac{z+1}{z-1}\right)^5 = 1$ (which is OK when $z$ isn't $1$: treat that case separately). $\endgroup$ – Daniel Freedman Dec 30 '11 at 1:46
  • $\begingroup$ Wow! I completely missed that. Thanks a lot! You can make your comment an answer if you want to. $\endgroup$ – E.O. Dec 30 '11 at 1:51
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    $\begingroup$ Alternatively, use the fact that $a^5-b^5 = (a-b)(a-\omega b)(a-\omega^2 b)(a-\omega^3 b)(a-\omega^4 b)$. $\endgroup$ – Álvaro Lozano-Robledo Dec 30 '11 at 1:53

Having calculated the fifth roots of unity $1, w, w^2, w^3, w^4$, i.e. the solutions to the equation $ u^5 = 1$, we can recast our equation as $ \left( \frac{z+1}{z-1} \right) ^5 = 1$ provided $z$ is not equal to $1$.

It is easy to verify that $z = 1$ is not a solution to the equation $(z+1)^5 = (z-1)^5$. So for $z$ not equal to $1$, we have that $ \frac{z+1}{z-1} = 1, w, w^2, w^3, w^4 $. Simple manipulation will allow you to express $z$ in terms of these roots.

Note that setting $ \frac{z+1}{z-1} = 1 $ will not yield a solution. This can be expected, since the equation we are solving is quartic and so has at most 4 distinct roots.


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