Find quintic roots of $x^ 5 = 1$

Question

How many quintic roots, i.e. fifth roots, does the number 1 have?

$$x^ 5 = 1$$

What are their values and how can one find them?

Answer 1

The equation $x^n=1$ has $n$ solutions, known as the $n$th roots of unity.

If $\zeta_n=e^{2\pi i/n}$, then the solutions are $\zeta_n^0,\zeta_n^1,\ldots,\zeta_n^{n-1}$.

These values, plotted on the complex plane, form a regular polygon for $n\geq3$ circumscribed by the unit circle.

For example, for $n=5$, the plot is presented below (from WolframAlpha)

Answer 2

If you are looking only for real roots, then the only real root is 1.

If you want the complex roots, then you want the five complex numbers that when raised to the fifth power are equal to 1.

We can consider that the number 1 can also be written as $e^{2k\pi i}$ for some integer k, and therefore: $$r_k^5=e^{2k\pi i}$$$$r_k=e^{\frac{2k\pi i}{5}}$$ We then consider the different possible values of k, noting that the result for $r_{k+5}=r_k$, and therefore we only need to plug in k=0,1,2,3,4. $$r_0=e^{\frac{0\pi i}{5}}=e^0=1$$$$r_1=e^{\frac{2\pi i}{5}}=\cos(\frac{2\pi}{5})+i\sin(\frac{2\pi}{5})$$$$r_2=e^{\frac{4\pi i}{5}}=\cos(\frac{4\pi}{5})+i\sin(\frac{4\pi}{5})$$$$r_3=e^{\frac{6\pi i}{5}}=\cos(\frac{6\pi}{5})+i\sin(\frac{6\pi}{5})$$$$r_4=e^{\frac{8\pi i}{5}}=\cos(\frac{8\pi}{5})+i\sin(\frac{8\pi}{5})$$

Answer 3

Note that $x^5=1$ can be factorized as

$$x^5-1=(x-1)\left(x^2+\frac{1+\sqrt5}2x+1\right)\left(x^2+\frac{1-\sqrt5}2x+1\right)=0$$

which leads to the roots

$$x=1,\> -\frac{1+\sqrt5}4\pm \frac i 4 \sqrt{10-2\sqrt5}, \> -\frac{1-\sqrt5}4\pm \frac i 4 \sqrt{10+2\sqrt5} $$