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Finding roots of unity using Euler and De Moivreś
June 14,2006 Compiled on June 27, 2025 at 7:41pm
1 Introduction
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.
2 Examples
2.1 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 \(\{1,-1\}\)
2.2 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 \(\{1,\) \(-\frac {1}{2}+i\frac {\sqrt {3}}{2},-\frac {1}{2}-i\frac {\sqrt {3}}{2}\}\)
2.3 Solve \(x-(-8)^{\frac {1}{3}}=0\)
\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*}
2.4 Find all values of \(\left ( -1\right ) ^{\frac {3}{4}}\)
Since \(-1=e^{i\pi }=\cos \left ( \pi \right ) +i\sin \pi \) then we have
\begin{align} \left ( -1\right ) ^{\frac {3}{4}} & =\left ( \cos \left ( \pi \right ) +i\sin \pi \right ) ^{\frac {3}{4}}\nonumber \\ & =\left [ \left ( \cos \left ( \pi \right ) +i\sin \pi \right ) ^{\frac {1}{4}}\right ] ^{3}\tag {1}\end{align}
Using \(\left ( \cos \alpha +i\sin \alpha \right ) ^{\frac {1}{n}}=\cos \left ( \frac {\alpha }{n}+k\frac {2\pi }{n}\right ) +i\sin \left ( \frac {\alpha }{n}+k\frac {2\pi }{n}\right ) \ \ \ \ \ \ \ k=0,1,\cdots n-1\) then (1) becomes
\[ \left ( \cos \left ( \pi \right ) +i\sin \pi \right ) ^{\frac {1}{4}}=\cos \left ( \frac {\pi }{n}+k\frac {2\pi }{n}\right ) +i\sin \left ( \frac {\pi }{n}+k\frac {2\pi }{n}\right ) \ \ \ \ \ \ \ k=0,1,\cdots n-1 \]
Hence, using \(n=4\)
\begin{align*} \left ( \cos \left ( \pi \right ) +i\sin \pi \right ) ^{\frac {1}{4}} & =\left [ \cos \left ( \frac {\pi }{4}\right ) +i\sin \left ( \frac {\pi }{4}\right ) ,\cos \left ( \frac {\pi }{4}+\frac {2\pi }{4}\right ) +i\sin \left ( \frac {\pi }{4}+\frac {2\pi }{4}\right ) ,\cos \left ( \frac {\pi }{4}+2\frac {2\pi }{4}\right ) +i\sin \left ( \frac {\pi }{4}+2\frac {2\pi }{4}\right ) ,\cos \left ( \frac {\pi }{n}+3\frac {2\pi }{4}\right ) +i\sin \left ( \frac {\pi }{4}+3\frac {2\pi }{4}\right ) \right ] \\ & =\left [ \cos \left ( \frac {\pi }{4}\right ) +i\sin \left ( \frac {\pi }{4}\right ) ,\cos \left ( \frac {\pi }{4}+\frac {\pi }{2}\right ) +i\sin \left ( \frac {\pi }{4}+\frac {\pi }{2}\right ) ,\cos \left ( \frac {\pi }{4}+\pi \right ) +i\sin \left ( \frac {\pi }{4}+\pi \right ) ,\cos \left ( \frac {\pi }{4}+\frac {3\pi }{2}\right ) +i\sin \left ( \frac {\pi }{4}+\frac {3\pi }{2}\right ) \right ] \\ & =\left [ \cos \left ( \frac {\pi }{4}\right ) +i\sin \left ( \frac {\pi }{4}\right ) ,\cos \left ( \frac {3}{4}\pi \right ) +i\sin \left ( \frac {3}{4}\pi \right ) ,\cos \left ( \frac {5}{4}\pi \right ) +i\sin \left ( \frac {5}{4}\pi \right ) ,\cos \left ( \frac {7}{4}\pi \right ) +i\sin \left ( \frac {7}{4}\pi \right ) \right ] \\ & =\left [ \left ( \frac {1}{2}+\frac {1}{2}i\right ) \sqrt {2},-\left ( \frac {1}{2}-\frac {1}{2}i\right ) \sqrt {2},-\left ( \frac {1}{2}+\frac {1}{2}i\right ) \sqrt {2},\left ( \frac {1}{2}-\frac {1}{2}i\right ) \sqrt {2}\right ] \end{align*}
Hence (1) becomes
\begin{align*} \left ( -1\right ) ^{\frac {3}{4}} & =\left [ \left ( \left ( \frac {1}{2}+\frac {1}{2}i\right ) \sqrt {2}\right ) ^{3},\left ( -\left ( \frac {1}{2}-\frac {1}{2}i\right ) \sqrt {2}\right ) ^{3},\left ( -\left ( \frac {1}{2}+\frac {1}{2}i\right ) \sqrt {2}\right ) ^{3},\left ( \left ( \frac {1}{2}-\frac {1}{2}i\right ) \sqrt {2}\right ) ^{3}\right ] \\ & =\left [ -\left ( \frac {1}{2}-\frac {1}{2}i\right ) \sqrt {2},\left ( \frac {1}{2}+\frac {1}{2}i\right ) \sqrt {2},\left ( \frac {1}{2}-\frac {1}{2}i\right ) \sqrt {2},-\left ( \frac {1}{2}+\frac {1}{2}i\right ) \sqrt {2}\right ] \\ & =\left [ -\left ( \frac {1-i}{2}\right ) \sqrt {2},\left ( \frac {1+i}{2}\right ) \sqrt {2},\left ( \frac {1-i}{2}\right ) \sqrt {2},-\left ( \frac {1+i}{2}\right ) \sqrt {2}\right ] \\ & =\left [ -\left ( \frac {1-i}{\sqrt {2}}\right ) ,\left ( \frac {1+i}{\sqrt {2}}\right ) ,\left ( \frac {1-i}{\sqrt {2}}\right ) ,-\left ( \frac {1+i}{\sqrt {2}}\right ) \right ] \end{align*}
The above are the 4 values of \(\allowbreak \left ( -1\right ) ^{\frac {3}{4}}\). In Mathematica the above can be found as follows
L=ResourceFunction["ComplexRoots"][(-1)^3,4]
ExpToTrig[L]