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While finding the 5th roots of unity $z^5=1$, I arrived at the following

$$(z-1)(z^4+z^3+z^2+z+1)=0$$

Now, I am well aware that I can arrive at the roots by using the fact that each root is separated by a value of $\alpha=2k\pi$

$$z^5=1 +0i \implies arg(z^5)=0$$ $$z^5=\cos(0 +2k\pi)+i\sin(0+2k\pi)$$ $$z=\cos(\frac{2k\pi}{5})+i\sin(\frac{2k\pi}{5})$$

Then, by letting $k$ take the values $0, ±1,±2$, I arrive at the roots

$$z=1,e^{i \frac{2\pi}{5}},e^{-i \frac{2\pi}{5}}, e^{i \frac{4\pi}{5}}, e^{-i\frac{4\pi}{5}}$$

However, for the sake of knowing if it is possible, is there a purely "algebraic" way of solving this quartic polynomial?

(Clearification: By "algebraic" I mean the classic ways you would deal with a regular polynomial such as the rational root theorem, factorization, or Newton-raphson etc)

insipidintegrator
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Filthyscrub
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1 Answers1

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Factorize the equation as follows \begin{align} z^5-1= &(z-1)(z^4+z^3+z^2+z+1)\\ =&(z-1)z^2(z^2+z+1+\frac1z+\frac1{z^2})\\ =&(z-1)z^2\left( (z+\frac1z)^2+(z+\frac1z)-1\right)\\ =&(z-1)z^2\left( z+\frac1z -\frac{1+\sqrt5}2\right)\left(z+\frac1z -\frac{1-\sqrt5}2\right)\\ =&(z-1)\left( z^2-\frac{1+\sqrt5}2z+1\right)\left(z^2 -\frac{1-\sqrt5}2 z+1\right)\\ \end{align} which yields the solutions $$z=1,\>\frac14\left(1+\sqrt5\pm i \sqrt{10-2\sqrt5}\right),\> \frac14\left(1-\sqrt5\pm i \sqrt{10+2\sqrt5}\right) $$

Quanto
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