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 \(n\).