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To find the roots of $$f(x)=x^n-1$$ Solving for $x$ from$$\begin{array}{rl}0& =x^n-1\\ x^n& =1\\ \text{(1)}& x& =1^{\frac{1}{n}}\end{array}$$

Now $1^{\frac{1}{n}}$ is evaluated. Since $$1=e^{i\left(2\pi \right)}$$ Substituting (2) in the RHS of (1) gives $$\begin{array}{rl}x& =(e^{i2\pi }{)}^{\frac{1}{n}}\\ \text{(3)}& & ={(\cos 2\pi +i\sin 2\pi )}^{\frac{1}{n}}\end{array}$$

Using De Moivre’s formula $${(\cos \alpha +i\sin \alpha )}^{\frac{1}{n}}=\cos (\frac{\alpha }{n}+k\frac{2\pi }{n})+i\sin (\frac{\alpha }{n}+k\frac{2\pi }{n})k=0,1,\cdots n-1$$ Therefore (3) is rewritten as$$x=\cos (\frac{2\pi }{n}+k\frac{2\pi }{n})+i\sin (\frac{2\pi }{n}+k\frac{2\pi }{n})k=0,1,\cdots n-1$$ 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 $k=0$ $$\begin{array}{rl}x& =\cos \left(\frac{2\pi }{2}\right)+i\sin \left(\frac{2\pi }{2}\right)\\ & =-1\end{array}$$

   And for $k=1$$$\begin{array}{rl}x& =\cos (\frac{2\pi }{2}+\frac{2\pi }{2})+i\sin (\frac{2\pi }{2}+\frac{2\pi }{2})\\ & =1\end{array}$$

   Hence the two roots are $\{1,-1\}$

2. Solve $f(x)=x^3-1$. Here $n=3$, hence for $k=0$$$\begin{array}{rl}x& =\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)\\ & =-\frac{1}{2}+i\frac{\sqrt{3}}{2}\end{array}$$

   And for $k=1$ $$\begin{array}{rl}x& =\cos (\frac{2\pi }{3}+\frac{2\pi }{3})+i\sin (\frac{2\pi }{3}+\frac{2\pi }{3})\\ & =\cos \left(\frac{4\pi }{3}\right)+i\sin \left(\frac{4\pi }{3}\right)\\ & =-\frac{1}{2}-i\frac{\sqrt{3}}{2}\end{array}$$

   And for $k=2$$$\begin{array}{rl}x& =\cos (\frac{2\pi }{3}+2\frac{2\pi }{3})+i\sin (\frac{2\pi }{3}+2\frac{2\pi }{3})\\ & =\cos \left(\frac{6\pi }{3}\right)+i\sin \left(\frac{6\pi }{3}\right)\\ & =1\end{array}$$

   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 $$\begin{array}{rl}x-(-8{)}^{\frac{1}{3}}& =0\\ x& =(-8{)}^{\frac{1}{3}}\\ & =(-8{)}^{\frac{1}{n}}\end{array}$$

Where $n=3$. But $8=8\left(1\right)=8e^{2\pi i}$. Hence the above becomes$$\begin{array}{rl}x& =(-8e^{2\pi i}{)}^{\frac{1}{n}}\\ & =-8^{\frac{1}{n}}{e}^{\frac{2\pi i}{n}}\\ \text{(1)}& & =-8^{\frac{1}{n}}{(\cos 2\pi +i\sin 2\pi )}^{\frac{1}{n}}\end{array}$$

But by De Moivre’s formula $${(\cos 2\pi +i\sin 2\pi )}^{\frac{1}{n}}=\cos (\frac{2\pi }{n}+k\frac{2\pi }{n})+i\sin (\frac{2\pi }{n}+k\frac{2\pi }{n})k=0\cdots n-1$$ Computing the above gives$${(\cos 2\pi +i\sin 2\pi )}^{\frac{1}{n}}=\{-\frac{1}{2}+i\frac{\sqrt{3}}{2},-\frac{1}{2}-i\frac{\sqrt{3}}{2},1\}$$ Hence from (1)$$\begin{array}{rl}x& =-8^{\frac{1}{3}}\{-\frac{1}{2}+i\frac{\sqrt{3}}{2},-\frac{1}{2}-i\frac{\sqrt{3}}{2},1\}\\ & =\{-8^{\frac{1}{3}}(-\frac{1}{2}+i\frac{\sqrt{3}}{2}),-8^{\frac{1}{3}}(-\frac{1}{2}-i\frac{\sqrt{3}}{2}),-8^{\frac{1}{3}}\}\\ & =\{1-1.732i,1+1.73i,-2\}\end{array}$$