Question

How do you use DeMoivre's theorem to solve ${{z}^{3}}-1=0$? \[\]

Answer 1

Hint: We recall Demoivre’s theorem from complex number with acute angle $\theta $ and exponent $n$ as ${{\left( \cos \theta +i\sin \theta \right)}^{n}}=\cos n\theta +i\sin n\theta $. We also recall that any complex number $z$ can be expressed as $z=r{{e}^{i\theta }}=r\left( \cos \theta +i\sin \theta \right)$. We put $z$ in the equation and see for what value of $\theta ,r$the equation ${{z}^{3}}-1=0$ satisfies. \[\]

Complete step by step answer:
We know any complex is commonly represented as $z=x+iy$ where $x$ is called the real part and $y$ is called the imaginary part. The complex number $z$ can also be represented in the polar form with $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ called modulus and $\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right),x>0$ as
\[z=r{{e}^{i\theta }}=r\left( \cos \theta +i\sin \theta \right)\]
We know from Demoivre’s theorem that for real angle $\theta $ and real integer $n$
\[{{\left( \cos \theta +i\sin \theta \right)}^{n}}=\cos n\theta +i\sin n\theta \]
We are asked to solve the following equation in the question
\[{{z}^{3}}-1=0\]
We see that we are given the complex variable $z$ with degree 3. We know that a polynomial of degree $n$ has $n$ roots. So we shall get three values of $z$. Let us consider
\[\begin{align}
  & {{z}^{3}}-1=0 \\
 & \Rightarrow {{z}^{3}}=1 \\
\end{align}\]
We put polar form of $z=r\left( \cos \theta +i\sin \theta \right)$in the above step to have;
\[\begin{align}
  & \Rightarrow {{\left( r\left( \cos \theta +i\sin \theta \right) \right)}^{3}}=1 \\
 & \Rightarrow {{r}^{3}}{{\left( \cos \theta +i\sin \theta \right)}^{3}}=1 \\
\end{align}\]
We use Demoivre’s theorem in the above step for $n=3$ to have
\[\begin{align}
  & \Rightarrow {{r}^{3}}\left( \cos 3\theta +i\sin 3\theta \right)=1 \\
 & \Rightarrow {{r}^{3}}\cos 3\theta +i{{r}^{3}}\sin 3\theta =1 \\
\end{align}\]
We compare the real part from both sides of the above equation to have ;
\[\Rightarrow {{r}^{3}}\cos 3\theta =1\]
Since modulus is always positive and real here the only possible value of $r=1$. So we have
\[\begin{align}
  & \Rightarrow \cos 3\theta =1 \\
 & \Rightarrow \cos 3\theta =\cos 2\pi \\
\end{align}\]
So the general solutions of the above trigonometric equation are
\[\begin{align}
  & \Rightarrow 3\theta =2k\pi \\
 & \Rightarrow \theta =\dfrac{2k\pi }{3} \\
\end{align}\]
Here $k$ is an arbitrary integer. So the solutions of the given equation are
\[\begin{align}
  & \text{if }k=0\text{ then }{{z}_{1}}=\cos \left( 0 \right)+i\sin \left( 0 \right)=1 \\
 & \text{if }k=1\text{ then }{{z}_{2}}=\cos \left( \dfrac{2\pi }{3} \right)+i\sin \left( \dfrac{2\pi }{3} \right)=\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \\
 & \text{if }k=0\text{ then }{{z}_{3}}=\cos \left( \dfrac{4\pi }{3} \right)+i\sin \left( \dfrac{4\pi }{3} \right)=\dfrac{-1}{2}-i\dfrac{\sqrt{3}}{2} \\
\end{align}\]

Note:
We note that we need to calculate for other values of $k$ since the period of cosine is $2\pi $ and all possible values of cosine occur in $\theta \in \left[ 0,2\pi \right)$. We note that the obtained solutions ${{z}_{1}},{{z}_{2}},{{z}_{3}}$are called cube roots of unity which have relation ${{z}_{1}}+{{z}_{2}}+{{z}_{3}}=0,{{z}_{1}}{{z}_{2}}{{z}_{3}}=1$. Similar to the solutions of the equation ${{z}^{n}}-1=0$ are $n,{{n}^{th}}$ roots of unity.