How can I give the solutions of $\ x^5-7$?

Question

I'm trying to solve the equation $\ x^5-7$, which have complex solutions, how can I calculate the solutions of the equation?

Thank you.

Answer 1

Do you mean $x^5-7 = 0$?

That would be a variation of the roots of unity.

$\sigma_1 = 7^{\frac 15}\\ \sigma_2 = 7^{\frac 15}(\cos \frac {2\pi}{5} + i\sin \frac {2\pi}{5})\\ \sigma_3 = 7^{\frac 15}(\cos \frac {4\pi}{5} + i\sin \frac {4\pi}{5})\\ \sigma_4 = 7^{\frac 15}(\cos \frac {6\pi}{5} + i\sin \frac {6\pi}{5})\\ \sigma_5 = 7^{\frac 15}(\cos \frac {8\pi}{5} + i\sin \frac {8\pi}{5})$

Answer 2

To expand on Doug's answer, you write $x=re^{i\theta}$ in general for $x\in\mathbb{C}$. The trivial root is $x=r^{1/n}$ where $r=7,n=5$ which is the only real-valued root. To find the angles in the complex plane, write $x=r^{1/n}e^{2\pi ik/n}$ with $k\in\{0,\ldots,n-1\}$.

Answer 3

Alternative hint (without polar forms and trig):  let $x = \sqrt[5]{7}\,z\,$, then the equation becomes:

$$ z^5 - 1 = 0 \;\;\iff\;\; (z-1)(z^4 + z^3+z^2+z+1) = 0 $$

This gives the root $z=1\,$, and the other $4$ roots can be found by solving the quartic, which reduces to a quadratic in $z+\cfrac{1}{z}\,$:

$$ z^2+z+1+\frac{1}{z}+\frac{1}{z^2}=0 \;\;\iff\;\; \left(z+\frac{1}{z}\right)^2+\left(z+\frac{1}{z}\right)-1=0 $$