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Question: How do you use DeMoivre's theorem to solve \({{z}^{3}}-1=0\)? \[\]...

How do you use DeMoivre's theorem to solve z3−1=0{{z}^{3}}-1=0? $$$$

Explanation

Solution

We recall Demoivre’s theorem from complex number with acute angle θ\theta and exponent nn as (cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ{{\left( \cos \theta +i\sin \theta \right)}^{n}}=\cos n\theta +i\sin n\theta . We also recall that any complex number zz can be expressed as z=reiθ=r(cos⁡θ+isin⁡θ)z=r{{e}^{i\theta }}=r\left( \cos \theta +i\sin \theta \right). We put zz in the equation and see for what value of θ,r\theta ,rthe equation z3−1=0{{z}^{3}}-1=0 satisfies. $$$$

Complete step by step answer:
We know any complex is commonly represented as z=x+iyz=x+iy where xx is called the real part and yy is called the imaginary part. The complex number zz can also be represented in the polar form with r=x2+y2r=\sqrt{{{x}^{2}}+{{y}^{2}}} called modulus and θ=tan⁡−1(yx),x>0\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right),x>0 as
z=reiθ=r(cos⁡θ+isin⁡θ)z=r{{e}^{i\theta }}=r\left( \cos \theta +i\sin \theta \right)
We know from Demoivre’s theorem that for real angle θ\theta and real integer nn
(cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ{{\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
z3−1=0{{z}^{3}}-1=0
We see that we are given the complex variable zz with degree 3. We know that a polynomial of degree nn has nn roots. So we shall get three values of zz. Let us consider

& {{z}^{3}}-1=0 \\\ & \Rightarrow {{z}^{3}}=1 \\\ \end{aligned}$$ We put polar form of $z=r\left( \cos \theta +i\sin \theta \right)$in the above step to have; $$\begin{aligned} & \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{aligned}$$ We use Demoivre’s theorem in the above step for $n=3$ to have $$\begin{aligned} & \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{aligned}$$ We compare the real part from both sides of the above equation to have ; $$\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{aligned} & \Rightarrow \cos 3\theta =1 \\\ & \Rightarrow \cos 3\theta =\cos 2\pi \\\ \end{aligned}$$ So the general solutions of the above trigonometric equation are $$\begin{aligned} & \Rightarrow 3\theta =2k\pi \\\ & \Rightarrow \theta =\dfrac{2k\pi }{3} \\\ \end{aligned}$$ Here $k$ is an arbitrary integer. So the solutions of the given equation are $$\begin{aligned} & \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{aligned}$$ **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.