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## Finding roots of unity using Euler and De Moivreś

June 14,2006   Compiled on January 30, 2024 at 4:57am

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.

1. 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 $$\{1,-1\}$$

2. 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*}