HOME

PDF (letter size)

We want to solve$$x^{\frac{n}{m}}=a$$ Where $n,m$ are integers. $n$ is called the power and $m$ is called the root. We start by writing the above as$${\left(x^{\frac{1}{m}}\right)}^n=a$$ Let $x^{\frac{1}{m}}=y$. The above becomes$$y^n=a$$ This is solved using De Moivre’s formula. $$\begin{array}{rl}y& =a^{\frac{1}{n}}\\ & ={(a\times 1)}^{\frac{1}{n}}\\ & ={\left(a{e}^{2i\pi }\right)}^{\frac{1}{n}}\end{array}$$

Since $1=e^{2\pi i}$. Using Euler formula $1=\cos \left(2\pi \right)+i\sin \left(2\pi \right)$. Hence$$y=a^{\frac{1}{n}}{(\cos \left(2\pi \right)+i\sin \left(2\pi \right))}^{\frac{1}{n}}$$ 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$$y=\{\begin{array}{c}{a}^{\frac{1}{3}}(\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right))\\ a^{\frac{1}{3}}(\cos (\frac{2\pi }{3}+\frac{2\pi }{3})+i\sin (\frac{2\pi }{3}+\frac{2\pi }{3}))\\ a^{\frac{1}{3}}(\cos (\frac{2\pi }{3}+\frac{4\pi }{3})+i\sin (\frac{2\pi }{3}+\frac{4\pi }{3}))\end{array}$$ Which simplifies to$$y=\{\begin{array}{c}{a}^{\frac{1}{3}}(\frac{1}{2}i\sqrt{3}-\frac{1}{2})\\ a^{\frac{1}{3}}(-\frac{1}{2}i\sqrt{3}-\frac{1}{2})\\ a^{\frac{1}{3}}\end{array}$$ Now we need to replace $y$ back to $x^{\frac{1}{m}}$ and the above becomes$$x^{\frac{1}{m}}=\{\begin{array}{c}{a}^{\frac{1}{3}}(\frac{1}{2}i\sqrt{3}-\frac{1}{2})\\ a^{\frac{1}{3}}(-\frac{1}{2}i\sqrt{3}-\frac{1}{2})\\ a^{\frac{1}{3}}\end{array}$$ Since the exponent now is a root, then$$x=\{\begin{array}{c}{\left(a^{\frac{1}{3}}(\frac{1}{2}i\sqrt{3}-\frac{1}{2})\right)}^m\\ {\left(a^{\frac{1}{3}}(-\frac{1}{2}i\sqrt{3}-\frac{1}{2})\right)}^m\\ a^{\frac{m}{3}}\end{array}$$ For example, if $m=2$$$x=\{\begin{array}{c}{\left(a^{\frac{1}{3}}(\frac{1}{2}i\sqrt{3}-\frac{1}{2})\right)}^2\\ {\left(a^{\frac{1}{3}}(-\frac{1}{2}i\sqrt{3}-\frac{1}{2})\right)}^2\\ a^{\frac{2}{3}}\end{array}$$ Notice that if the solution $x$ is meant to be real, then the above reduces to$$x=a^{\frac{2}{3}}$$ And for $m=4$$$\begin{array}{rl}x& =\{\begin{array}{c}{a}^{\frac{4}{3}}{(\frac{1}{2}i\sqrt{3}-\frac{1}{2})}^4\\ a^{\frac{4}{3}}{(-\frac{1}{2}i\sqrt{3}-\frac{1}{2})}^4\\ a^{\frac{4}{3}}\end{array}\\ & =\{\begin{array}{c}{a}^{\frac{4}{3}}(-\frac{1}{2}+\frac{i\sqrt{3}}{2})\\ a^{\frac{4}{3}}(-\frac{1}{2}-\frac{i\sqrt{3}}{2})\\ a^{\frac{4}{3}}\end{array}\end{array}$$

Notice that if the solution $x$ is meant to be real, then the above reduces to$$x=a^{\frac{4}{3}}$$ For $a\ge 0$.  And so on. For the case of power $n$ being negative integer, for example,$$x^{\frac{-3}{2}}=a$$ 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 $n$.