To find the roots of \[ f(x)=x^{n}-1 \] Solving for \(x\) from\begin {align} 0 & =x^{n}-1\nonumber \\ x^{n} & =1\nonumber \\ x & =1^{\frac {1}{n}}\tag {1} \end {align}
Now \(1^{\frac {1}{n}}\) is evaluated. Since \[ 1=e^{i\left ( 2\pi \right ) }\] Substituting (2) in the RHS of (1) gives \begin {align} x & =(e^{i2\pi })^{\frac {1}{n}}\nonumber \\ & =\left ( \cos 2\pi +i\sin 2\pi \right ) ^{\frac {1}{n}}\tag {3} \end {align}
Using De Moivre’s formula \[ \left ( \cos \alpha +i\sin \alpha \right ) ^{\frac {1}{n}}=\cos \left ( \frac {\alpha }{n}+k\frac {2\pi }{n}\right ) +i\sin \left ( \frac {\alpha }{n}+k\frac {2\pi }{n}\right ) \ \ \ \ \ \ \ k=0,1,\cdots n-1 \] Therefore (3) is rewritten as\[ x=\cos \left ( \frac {2\pi }{n}+k\frac {2\pi }{n}\right ) +i\sin \left ( \frac {2\pi }{n}+k\frac {2\pi }{n}\right ) \ \ \ \ \ \ \ k=0,1,\cdots n-1 \] The above gives the roots of \(f(x)=x^{n}-1\). The following examples illustrate the use of the above.
Solve \(f(x)=x^{2}-1\). Here \(n=2\), therefore \(k=0,1\). For \(k=0\) \begin {align*} x & =\cos \left ( \frac {2\pi }{2}\right ) +i\sin \left ( \frac {2\pi }{2}\right ) \\ & =-1 \end {align*}
And for \(k=1\)\begin {align*} x & =\cos \left ( \frac {2\pi }{2}+\frac {2\pi }{2}\right ) +i\sin \left ( \frac {2\pi }{2}+\frac {2\pi }{2}\right ) \\ & =1 \end {align*}
Hence the two roots are \(\{1,-1\}\)
Solve \(f(x)=x^{3}-1\). Here \(n=3\), hence for \(k=0\)\begin {align*} x & =\cos \left ( \frac {2\pi }{3}\right ) +i\sin \left ( \frac {2\pi }{3}\right ) \\ & =-\frac {1}{2}+i\frac {\sqrt {3}}{2} \end {align*}
And for \(k=1\) \begin {align*} x & =\cos \left ( \frac {2\pi }{3}+\frac {2\pi }{3}\right ) +i\sin \left ( \frac {2\pi }{3}+\frac {2\pi }{3}\right ) \\ & =\cos \left ( \frac {4\pi }{3}\right ) +i\sin \left ( \frac {4\pi }{3}\right ) \\ & =-\frac {1}{2}-i\frac {\sqrt {3}}{2} \end {align*}
And for \(k=2\)\begin {align*} x & =\cos \left ( \frac {2\pi }{3}+2\frac {2\pi }{3}\right ) +i\sin \left ( \frac {2\pi }{3}+2\frac {2\pi }{3}\right ) \\ & =\cos \left ( \frac {6\pi }{3}\right ) +i\sin \left ( \frac {6\pi }{3}\right ) \\ & =1 \end {align*}
Therefore the roots are \(\{1,\) \(-\frac {1}{2}+i\frac {\sqrt {3}}{2},-\frac {1}{2}-i\frac {\sqrt {3}}{2}\}\)
Here is another example. Let us solve \begin {align*} x-(-8)^{\frac {1}{3}} & =0\\ x & =(-8)^{\frac {1}{3}}\\ & =(-8)^{\frac {1}{n}} \end {align*}
Where \(n=3\). But \(8=8\left ( 1\right ) =8e^{2\pi i}\). Hence the above becomes\begin {align} x & =(-8e^{2\pi i})^{\frac {1}{n}}\nonumber \\ & =-8^{\frac {1}{n}}e^{\frac {2\pi i}{n}}\nonumber \\ & =-8^{\frac {1}{n}}\left ( \cos 2\pi +i\sin 2\pi \right ) ^{\frac {1}{n}}\tag {1} \end {align}
But by De Moivre’s formula \[ \left ( \cos 2\pi +i\sin 2\pi \right ) ^{\frac {1}{n}}=\cos \left ( \frac {2\pi }{n}+k\frac {2\pi }{n}\right ) +i\sin \left ( \frac {2\pi }{n}+k\frac {2\pi }{n}\right ) \qquad k=0\cdots n-1 \] Computing the above gives\[ \left ( \cos 2\pi +i\sin 2\pi \right ) ^{\frac {1}{n}}=\left \{ -\frac {1}{2}+i\frac {\sqrt {3}}{2},-\frac {1}{2}-i\frac {\sqrt {3}}{2},1\right \} \] Hence from (1)\begin {align*} x & =-8^{\frac {1}{3}}\left \{ -\frac {1}{2}+i\frac {\sqrt {3}}{2},-\frac {1}{2}-i\frac {\sqrt {3}}{2},1\right \} \\ & =\left \{ -8^{\frac {1}{3}}\left ( -\frac {1}{2}+i\frac {\sqrt {3}}{2}\right ) ,-8^{\frac {1}{3}}\left ( -\frac {1}{2}-i\frac {\sqrt {3}}{2}\right ) ,-8^{\frac {1}{3}}\right \} \\ & =\left \{ 1-1.732i,1+1.73i,-2\right \} \end {align*}