Question

Prove that $n$, ${{n}^{\text{th}}}$ roots of unity are in GP. $$$$

Answer 1

We recall the geometric progression where there is a common ratio between consecutive terms. We express of ${{n}^{\text{th}}}$ roots of unity in the form ${{\left( \cos 2k\pi +i\sin 2k\pi \right)}^{\dfrac{1}{n}}}$ and use the identity ${{\left( \cos \theta +i\sin \theta \right)}^{m}}=\cos m\theta +i\sin m\theta ,m\in R$. We put different values of $k$ to find different roots of the unit and see if they are in GP or not.

Complete step-by-step solution:
We know that a geometric sequence otherwise known as Geometric progression, abbreviated as GP is a type sequence where the ratio between any two consecutive numbers is constant . If $\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...$ is an GP, then
$\dfrac{{{x}_{2}}}{{{x}_{1}}}=\dfrac{{{x}_{3}}}{{{x}_{1}}}...=r...(1)$
We know from complex numbers that all complex numbers can be expressed as $r{{e}^{i\theta }}$ where
${{e}^{i\theta }}=\cos \theta +i\sin \theta$
We also know the identity
${{\left( \cos \theta +i\sin \theta \right)}^{m}}=\cos m\theta +i\sin m\theta ,m\in R$
We know we shall get $n$ roots of an polynomial equation if the highest degree on variable is $n$. So we shall get $n$ roots of the equation ${{x}^{n}}=1$ and those roots are roots of unity. Let us consider
We put $\cos 2k\pi$ in place of 1 and $i\sin 2k\pi$ in place of 0 for some integer $k$ since we know $\cos 2k\pi =1,\sin 2k\pi =0$ for all $k\in Z$ to have;
$\Rightarrow x={{\left( \cos 2k\pi +i\sin 2k\pi \right)}^{\dfrac{1}{n}}}$
We use the identity ${{\left( \cos \theta +i\sin \theta \right)}^{m}}=\cos m\theta +i\sin m\theta ,m\in R$ for $\theta =2k\pi$ and $m=\dfrac{1}{n}$ in the above step to have;
$\Rightarrow x=\cos \dfrac{2k\pi }{n}+i\sin \dfrac{2k\pi }{n}$
The above expression is an expression for ${{n}^{\text{th}}}$ roots of unity. Since $k$ is an integer and $n$ is also an integer the possible values of $k$ will be less than $n$ which are $k=0,1,2,...,n-1$. So we put $k=0,1,2,...,n-1$ in the express of $x$ to have to

& \Rightarrow x=\cos \dfrac{2\left( 0 \right)\pi }{n}+i\sin \dfrac{2\left( 0 \right)\pi }{n}=1 \\\ & \Rightarrow x=\cos \dfrac{2\left( 1 \right)\pi }{n}+i\sin \dfrac{2\left( 1 \right)\pi }{n}=\cos \dfrac{2\pi }{n}+i\sin \dfrac{2\pi }{n} \\\ & \Rightarrow x=\cos \dfrac{2\left( 2 \right)\pi }{n}+i\sin \dfrac{2\left( 2 \right)\pi }{n}=\cos \dfrac{4\pi }{n}+i\sin \dfrac{4\pi }{n} \\\ & \Rightarrow x=\cos \dfrac{2\left( 3 \right)\pi }{n}+i\sin \dfrac{2\left( 3 \right)\pi }{n}=\cos \dfrac{6\pi }{n}+i\sin \dfrac{6\pi }{n} \\\ & \vdots \\\ & \Rightarrow x=\cos \dfrac{2\left( n-1 \right)\pi }{n}+i\sin \dfrac{2\left( n-1 \right)\pi }{n} \\\ \end{aligned}$$ Let us denote $\alpha =\cos \dfrac{2\pi }{n}+i\sin \dfrac{2\pi }{n}$. We use the identity ${{\left( \cos \theta +i\sin \theta \right)}^{m}}=\cos m\theta +i\sin m\theta $ for $\theta =2\pi $ and $m=2$ to find ${{\alpha }^{2}}$ as; $${{\alpha }^{2}}={{\left( \cos \dfrac{2\pi }{n}+i\sin \dfrac{2\pi }{n} \right)}^{2}}=\cos \dfrac{4\pi }{n}+i\sin \dfrac{4\pi }{n}$$ We can similarly find ${{\alpha }^{3}}$ as $${{\alpha }^{2}}={{\left( \cos \dfrac{2\pi }{n}+i\sin \dfrac{2\pi }{n} \right)}^{3}}=\cos \dfrac{6\pi }{n}+i\sin \dfrac{6\pi }{n}$$ We can go on to express all the $n$ roots of unity as an exponent until ${{\alpha }^{n-1}}$ which we find below, $${{\alpha }^{2}}={{\left( \cos \dfrac{2\pi }{n}+i\sin \dfrac{2\pi }{n} \right)}^{n-1}}=\cos \dfrac{2\left( n-1 \right)\pi }{n}+i\sin \dfrac{2\left( n-1 \right)\pi }{n}$$ So we can write all the $n$ roots of unity as; $$1,\alpha ,{{\alpha }^{2}},...,{{\alpha }^{n-1}}$$ We see that the ratio between any two terms is $\dfrac{\alpha }{1}=\dfrac{{{\alpha }^{2}}}{\alpha }=...=\dfrac{{{\alpha }^{n-1}}}{{{\alpha }^{n-2}}}=\alpha $. Hence the $n$, ${{n}^{\text{th}}}$ roots of unity are in GP. Hence it is proved. $$$$ **Note:** We note that the ${{n}^{\text{th}}}$ roots of unity can be expressed in the form ${{e}^{i\dfrac{2k\pi }{n}}}=\cos \dfrac{2k\pi }{n}+i\sin \dfrac{2k\pi }{n}$. All the $n$ roots of unity lie on a circle with centre as the origin and radius of 1 unit in the complex plane. The sum of $n,{{n}^{\text{th}}}$ roots of unity is zero which means $1+{{\alpha }^{2}}+...+{{\alpha }^{n-1}}=0$ and product of $n,{{n}^{\text{th}}}$ roots of unity is 1 which means $1\times {{\alpha }^{2}}\times ...\times {{\alpha }^{n-1}}=0$.