<s>[INST] Complex roots of quartic equation during solution of $z^5=1$ 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)
 [/INST] 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
$$z=1,\>\frac14\left(1+\sqrt5\pm i \sqrt{10-2\sqrt5}\right),\> 
 \frac14\left(1-\sqrt5\pm i \sqrt{10+2\sqrt5}\right)
$$
</s>