Small note on solving xnm=a

We want to solve

Where $n,m$ are integers. $n$ is called the power and $m$ is called the root. We start by writing the above as

Let $x^{\frac{1}{m}}=y$. The above becomes

This is solved using De Moivre’s formula.

	$y$	$\phantom{}=a^{\frac{1}{n}}$
		$\phantom{}={\left(a\times 1\right)}^{\frac{1}{n}}$
		$\phantom{}={\left(a{e}^{2i\pi }\right)}^{\frac{1}{n}}$

Since $1=e^{2\pi i}$. Using Euler formula $1=\cos \left(2\pi \right)+i\sin \left(2\pi \right)$. Hence

$$y=a^{\frac{1}{n}}{\left(\cos \left(2\pi \right)+i\sin \left(2\pi \right)\right)}^{\frac{1}{n}}$$

But by De Moivre’s formula

$${(\cos \left(2\pi \right)+i\sin \left(2\pi \right))}^{\frac{1}{n}}=\cos (\frac{2\pi }{n}+k\frac{2\pi }{n})+i\sin (\frac{2\pi }{n}+k\frac{2\pi }{n})k=0,1,\cdots n-1$$

Therefore

$$y=a^{\frac{1}{n}}(\cos (\frac{2\pi }{n}+k\frac{2\pi }{n})+i\sin (\frac{2\pi }{n}+k\frac{2\pi }{n}))k=0,1,\cdots n-1$$

For example, let $n=3$ then we have 3 solutions

$$y=\{\begin{array}{c}{a}^{\frac{1}{3}}\left(\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)\right)\\ a^{\frac{1}{3}}\left(\cos \left(\frac{2\pi }{3}+\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}+\frac{2\pi }{3}\right)\right)\\ a^{\frac{1}{3}}\left(\cos \left(\frac{2\pi }{3}+\frac{4\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}+\frac{4\pi }{3}\right)\right)\end{array}\phantom{}$$

Which simplifies to

$$y=\{\begin{array}{c}{a}^{\frac{1}{3}}\left(\frac{1}{2}i\sqrt{3}-\frac{1}{2}\right)\\ a^{\frac{1}{3}}\left(-\frac{1}{2}i\sqrt{3}-\frac{1}{2}\right)\\ a^{\frac{1}{3}}\end{array}\phantom{}$$

Now we need to replace $y$ back to $x^{\frac{1}{m}}$ and the above becomes

$$x^{\frac{1}{m}}=\{\begin{array}{c}{a}^{\frac{1}{3}}\left(\frac{1}{2}i\sqrt{3}-\frac{1}{2}\right)\\ a^{\frac{1}{3}}\left(-\frac{1}{2}i\sqrt{3}-\frac{1}{2}\right)\\ a^{\frac{1}{3}}\end{array}\phantom{}$$

Since the exponent now is a root, then

$$x=\{\begin{array}{c}{\left(a^{\frac{1}{3}}\left(\frac{1}{2}i\sqrt{3}-\frac{1}{2}\right)\right)}^m\\ {\left(a^{\frac{1}{3}}\left(-\frac{1}{2}i\sqrt{3}-\frac{1}{2}\right)\right)}^m\\ a^{\frac{m}{3}}\end{array}\phantom{}$$

For example, if $m=2$

$$x=\{\begin{array}{c}{\left(a^{\frac{1}{3}}\left(\frac{1}{2}i\sqrt{3}-\frac{1}{2}\right)\right)}^2\\ {\left(a^{\frac{1}{3}}\left(-\frac{1}{2}i\sqrt{3}-\frac{1}{2}\right)\right)}^2\\ a^{\frac{2}{3}}\end{array}\phantom{}$$

Notice that if the solution $x$ is meant to be real, then the above reduces to

And for $m=4$

	$x$	$\phantom{}=\{\begin{array}{c}{a}^{\frac{4}{3}}{\left(\frac{1}{2}i\sqrt{3}-\frac{1}{2}\right)}^4\\ a^{\frac{4}{3}}{\left(-\frac{1}{2}i\sqrt{3}-\frac{1}{2}\right)}^4\\ a^{\frac{4}{3}}\end{array}\phantom{}$
		$\phantom{}=\{\begin{array}{c}{a}^{\frac{4}{3}}\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)\\ a^{\frac{4}{3}}\left(-\frac{1}{2}-\frac{i\sqrt{3}}{2}\right)\\ a^{\frac{4}{3}}\end{array}\phantom{}$

Notice that if the solution $x$ is meant to be real, then the above reduces to

For $a\ge 0$. And so on. For the case of power $n$ being negative integer, for example,

Then let $n=3$ and move the negative sign to the denominator to become $x^{\frac{3}{-2}}$. This way we can now use De Moivre’s formula for positive $n$.