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## Finding roots of unity using Euler and De Moivreś

June 14,2006   Compiled on May 22, 2020 at 4:22am

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 $$\{1,$$ $$-\frac{1}{2}+i\frac{\sqrt{3}}{2},-\frac{1}{2}-i\frac{\sqrt{3}}{2}\}$$