How would you find the roots of the following equation?

Question

How would you find the roots of the following equation?

$$x^{13} + 1 = 0.$$

I am absolutely new at this and have no clue.

Answer 1

$$x^{13}+1=0$$

$$x^{13}=-1$$

$$x=(-1)^{\frac 1{13}}$$

$$ x= (\cos \pi+i\sin \pi )^{\frac 1{13}}$$ Then by De Moivre's formula, add $2k\pi$ with the arguments, and then you can use as

$$ x= (\cos (2k\pi+\pi)+i\sin (2k\pi+\pi) )^{\frac 1{13}}$$ $$ x= \cos \frac{2k\pi+\pi}{13}+i\sin \frac{2k\pi+\pi}{13}=e^{\frac{2k+1}{13}\pi}$$

Now substitute, $k=0,1,2,\ldots 12$ which will give you the required 13 solutions of the given equation.

$$\text{for } k=0,\quad x= e^{\pi/13}$$ $$\text{for } k=1,\quad x= e^{3\pi/13}$$ $$\vdots$$ $$\text{for } k=12,\quad x= e^{25\pi/13}$$

Answer 2

Rearranging, we have that $x^{13} = -1.$ An obvious solution is $x = -1.$ To find the others, we take a look at the complex plane. The $13$ roots can be shown to be spaced evenly around the unit circle. We already know one solution, so the others follow as such: $\boxed{x = -1, i\sin(\frac{\pi}{13}) - \cos(\frac{\pi}{13}), i\sin(\frac{2\pi}{13}) - \cos(\frac{2\pi}{13}),..., i\sin(\frac{12\pi}{13}) - \cos(\frac{12\pi}{13})}.$