PDF (letter size)

## Small note on solving $$x^{\frac {n}{m}}=a$$

March 25, 2024   Compiled on March 25, 2024 at 6:44am

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 {align*} y & =a^{\frac {1}{n}}\\ & =\left ( a\times 1\right ) ^{\frac {1}{n}}\\ & =\left ( ae^{2i\pi }\right ) ^{\frac {1}{n}} \end {align*}

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$\left ( \cos \left ( 2\pi \right ) +i\sin \left ( 2\pi \right ) \right ) ^{\frac {1}{n}}=\cos \left ( \frac {2\pi }{n}+k\frac {2\pi }{n}\right ) +i\sin \left ( \frac {2\pi }{n}+k\frac {2\pi }{n}\right ) \qquad k=0,1,\cdots n-1$ Therefore $y=a^{\frac {1}{n}}\left ( \cos \left ( \frac {2\pi }{n}+k\frac {2\pi }{n}\right ) +i\sin \left ( \frac {2\pi }{n}+k\frac {2\pi }{n}\right ) \right ) \qquad k=0,1,\cdots n-1$ For example, let $$n=3$$ then we have 3 solutions$y=\left \{ \begin {array} [c]{c}a^{\frac {1}{3}}\left ( \cos \left ( \frac {2\pi }{3}\right ) +i\sin \left ( \frac {2\pi }{3}\right ) \right ) \\ a^{\frac {1}{3}}\left ( \cos \left ( \frac {2\pi }{3}+\frac {2\pi }{3}\right ) +i\sin \left ( \frac {2\pi }{3}+\frac {2\pi }{3}\right ) \right ) \\ a^{\frac {1}{3}}\left ( \cos \left ( \frac {2\pi }{3}+\frac {4\pi }{3}\right ) +i\sin \left ( \frac {2\pi }{3}+\frac {4\pi }{3}\right ) \right ) \end {array} \right .$ Which simpliﬁes to$y=\left \{ \begin {array} [c]{c}a^{\frac {1}{3}}\left ( \frac {1}{2}i\sqrt {3}-\frac {1}{2}\right ) \\ a^{\frac {1}{3}}\left ( -\frac {1}{2}i\sqrt {3}-\frac {1}{2}\right ) \\ a^{\frac {1}{3}}\end {array} \right .$ Now we need to replace $$y$$ back to $$x^{\frac {1}{m}}$$ and the above becomes$x^{\frac {1}{m}}=\left \{ \begin {array} [c]{c}a^{\frac {1}{3}}\left ( \frac {1}{2}i\sqrt {3}-\frac {1}{2}\right ) \\ a^{\frac {1}{3}}\left ( -\frac {1}{2}i\sqrt {3}-\frac {1}{2}\right ) \\ a^{\frac {1}{3}}\end {array} \right .$ Since the exponent now is a root, then$x=\left \{ \begin {array} [c]{c}\left ( a^{\frac {1}{3}}\left ( \frac {1}{2}i\sqrt {3}-\frac {1}{2}\right ) \right ) ^{m}\\ \left ( a^{\frac {1}{3}}\left ( -\frac {1}{2}i\sqrt {3}-\frac {1}{2}\right ) \right ) ^{m}\\ a^{\frac {m}{3}}\end {array} \right .$ For example, if $$m=2$$$x=\left \{ \begin {array} [c]{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} \right .$ 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 {align*} x & =\left \{ \begin {array} [c]{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} \right . \\ & =\left \{ \begin {array} [c]{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} \right . \end {align*}

Notice that if the solution $$x$$ is meant to be real, then the above reduces to$x=a^{\frac {4}{3}}$ For $$a\geq 0$$.  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 $$n$$.