Question

Let $\alpha =\dfrac{\cos 2\pi }{n}+i\dfrac{\sin 2\pi }{n}$ where $n\in N,n>2$. Prove that for the complex numbers ${{z}_{1}},{{z}_{2}}$ , $\sum\limits_{r=0}^{n-1}{{{\left| {{z}_{1}}+{{\alpha }^{r}}{{z}_{2}} \right|}^{2}}}=n{{\left| {{z}_{1}} \right|}^{2}}$\[\]

Answer 1

Hint: The given expression $\alpha =\dfrac{\cos 2\pi }{n}+i\dfrac{\sin 2\pi }{n}$ is the expression for ${{n}^{\text{th}}}$ roots of unity. All the roots of unity with polynomial of degree $n$ are related by ${{\alpha }^{n}}=1$ and $1+\alpha +{{\alpha }^{2}}+...+{{\alpha }^{n-1}}=0$ . We use these relations when we apply the method induction to prove. We first prove the statement for $n=3$, then assume for $n=k$ and the prove again for $n=k+1.$\[\]

Complete step by step answer:
We know that any complex number $z=x+iy$ where $x$and $y$ are real numbers . The modulus of a complex number $z$ is the distance from the origin in the complex plane which is defined as $\left| z \right|=\left| x\pm iy \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$. We know that ${{\left| z \right|}^{2}}=z\overline{z}$. We are asked to prove the statement ,
\[\sum\limits_{r=0}^{n-1}{{{\left| {{z}_{1}}+{{\alpha }^{r}}{{z}_{2}} \right|}^{2}}}=n\left( {{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}} \right)\]
We are given two complex numbers ${{z}_{1}},{{z}_{2}}$. We are given a complex number in Euler’s co-ordinates. We have for the natural number $n>2$,
\[\alpha =\dfrac{\cos 2\pi }{n}+i\dfrac{\sin 2\pi }{n}\]
We know that above expression the expression for ${{n}^{\text{th}}}$ roots of unit which means a solution of polynomial of equation ${{x}^{n}}=1$. We also know that ${{\alpha }^{n}}=1$ and $1+\alpha +{{\alpha }^{2}}+...+{{\alpha }^{n-1}}=0$. let us find ${{\left| {{z}_{1}}+{{\alpha }^{r}}{{z}_{2}} \right|}^{2}}$ before we proceed. So we have
\[\begin{align}
  & {{\left| {{z}_{1}}+{{\alpha }^{r}}{{z}_{2}} \right|}^{2}} \\
 & =\left( {{z}_{1}}+{{\alpha }^{r}}{{z}_{2}} \right)\left( \overline{{{z}_{1}}}+{{\alpha }^{r}}\overline{{{z}_{2}}} \right) \\
 & ={{z}_{1}}\overline{{{z}_{1}}}+{{\alpha }^{2r}}{{z}_{2}}\overline{{{z}_{2}}}+{{\alpha }^{r}}\left( {{z}_{2}}\overline{{{z}_{1}}}+{{z}_{1}}\overline{{{z}_{2}}} \right) \\
 & ={{\left| {{z}_{1}} \right|}^{2}}+{{\alpha }^{2r}}{{\left| {{z}_{2}} \right|}^{2}}+{{\alpha }^{r}}\left( {{z}_{2}}\overline{{{z}_{1}}}+{{z}_{1}}\overline{{{z}_{2}}} \right) \\
\end{align}\]
We shall prove the statement the statement by induction. So we take prove for the first value. Let us expand the left hand side of the statement $n=3$ . When we take $n=3$ we shall get the three cube roots unity as ${{\alpha }^{0}}=1,{{\alpha }^{1}}=\alpha ,{{\alpha }^{2}}$ then we have$1+\alpha +{{\alpha }^{2}}=0,{{\alpha }^{4}}={{\alpha }^{3}}\times \alpha =1\times \alpha =\alpha $. Now we use it and proceed ,
\[\begin{align}
  & \sum\limits_{r=0}^{3-1}{{{\left| {{z}_{1}}+{{\alpha }^{r}}{{z}_{2}} \right|}^{2}}}=\sum\limits_{r=0}^{2}{{{\left| {{z}_{1}}+{{\alpha }^{r}}{{z}_{2}} \right|}^{2}}} \\
 & ={{\left| {{z}_{1}}+{{\alpha }^{0}}{{z}_{2}} \right|}^{2}}+{{\left| {{z}_{1}}+{{\alpha }^{1}}{{z}_{2}} \right|}^{2}}+{{\left| {{z}_{1}}+{{\alpha }^{2}}{{z}_{2}} \right|}^{2}} \\
 & =3{{\left| {{z}_{1}} \right|}^{2}}+\left( {{\alpha }^{2\times 0}}+{{\alpha }^{2\times 1}}+{{\alpha }^{2\times 2}} \right){{\left| {{z}_{2}} \right|}^{2}}+\left( {{z}_{2}}\overline{{{z}_{1}}}+{{z}_{1}}\overline{{{z}_{2}}} \right)\left( {{\alpha }^{0}}+{{\alpha }^{1}}+{{\alpha }^{2}} \right) \\
 & =3{{\left| {{z}_{1}} \right|}^{2}} \\
\end{align}\]
So the statement is true for $n=3$. Now we shall assume the statement is true for $n=k$ . So we have
\[\begin{align}
  & \sum\limits_{r=0}^{k-1}{{{\left| {{z}_{1}}+{{\alpha }^{r}}{{z}_{2}} \right|}^{2}}}=k{{\left| {{z}_{1}} \right|}^{2}} \\
 & \Rightarrow {{\left| {{z}_{1}}+{{\alpha }^{0}}{{z}_{2}} \right|}^{2}}+{{\left| {{z}_{1}}+{{\alpha }^{1}}{{z}_{2}} \right|}^{2}}+...+{{\left| {{z}_{1}}+{{\alpha }^{k-1}}{{z}_{2}} \right|}^{2}}=\left( k-1 \right){{\left| {{z}_{1}} \right|}^{2}} \\
\end{align}\]
Now we have to prove for $n=k+1$ which means we have to prove ${{\sum\limits_{r=0}^{k}{\left| {{z}_{1}}+{{\alpha }^{r}}{{z}_{2}} \right|}}^{2}}=n{{\left| {{z}_{1}} \right|}^{2}}$. We have the ${{\left( k+1 \right)}^{\text{th}}}$ root as ${{\alpha }^{k}}=0$. We write the left hand side of the statement for $n=k+1$ and expand using truth of the statement for $n=k$ ,
 \[\begin{align}
  & \sum\limits_{r=0}^{k}{{{\left| {{z}_{1}}+{{\alpha }^{r}}{{z}_{2}} \right|}^{2}}} \\
 & ={{\left| {{z}_{1}}+{{\alpha }^{0}}{{z}_{2}} \right|}^{2}}+{{\left| {{z}_{1}}+{{\alpha }^{1}}{{z}_{2}} \right|}^{2}}+...+{{\left| {{z}_{1}}+{{\alpha }^{k-1}}{{z}_{2}} \right|}^{2}}+{{\left| {{z}_{1}}+{{\alpha }^{k}}{{z}_{2}} \right|}^{2}} \\
 & =\left( k-1 \right){{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{1}} \right|}^{2}}\left( \because {{\alpha }^{k}}=0 \right) \\
 & =k{{\left| {{z}_{1}} \right|}^{2}} \\
\end{align}\]
Now we have the statement is true for $n=3$ and if assume the statement is true for $n=k$ it implies that the statement is true for $n=k+1$. So by induction the statement is true for all $n\in N.$ Hence proved.\[\]

Note: We note that a complex number can also be written in Euler’s form as $z=r\cos \theta +ir\sin \theta =r{{e}^{i\theta }}$ where $r$ is the modulus and $\theta $ is the angle $z$ makes when joined with origin, otherwise known as argument $\theta =\arg \left( z \right)$. So the given $\alpha $ can be written $\alpha ={{e}^{i\dfrac{2\pi }{n}}}$.