The calculation of roots of complex numbers.

Question

How to calculate the roots of $x^6+64=0$?

Or how to calculate the roots of $1+x^{2n}=0$?

Give its easy and understanble solution method. Thank you. In general, the results of "exp" are obtained.

Answer 1

$$x^{2n}=-1=\cos\pi+i\sin\pi=\cos(2k+1)\pi+i\sin(2k+1)\pi$$ where $k$ is any integer

Using De Moivre's formula,

$$\implies x=\cos\left(\frac{(2k+1)\pi}{2n}\right)+i\sin\left(\frac{(2k+1)\pi}{2n}\right)$$ where $k=0,1,2\cdots, 2n-1$ as $2n$ degree equation has exactly $2n$ roots

In fact, we can make some further generalization on the set of values of $k,$ see here

Answer 2

$$x^6+64=(x^2)^3+(2^2)^3$$

This means that $x^2+2^2$ is one factor of $x^6+64$, and so two of the solutions are $x=\pm2i$.

Then we have the remaining factors:

$$x^6+64=(x^2+2^2)(x^4-2^2x^2+2^4)$$

and the quadratic result is

$$x^2={4\pm\sqrt{16-64}\over 2}=2\pm2\sqrt{-3}$$

This produces two solutions $x^2=2\pm2i\sqrt 3$, and each solution has two further solutions for $x$.