Question

Given $ z=\cos \left( \dfrac{2\pi }{2n+1} \right)+i\sin \left( \dfrac{2\pi }{2n+1} \right) $ where $ n $ a positive integer is find the equation whose roots are $ \alpha =z+{{z}^{3}}+...+{{z}^{2n-1}} $ and $ \beta ={{z}^{2}}+{{z}^{4}}+...+{{z}^{2n}} $ .

Answer 1

Hint: We use the identity $ {{\left( \cos \theta +i\sin \theta \right)}^{m}}=\cos m\theta +i\sin m\theta $ to find $ {{z}^{2n+1}} $ . We find $ \alpha +\beta $ using obtained $ {{z}^{2n+1}} $ and sum of $ n $ sum of $ n $ terms of a geometric progression(GP). We use sum of $ n $ terms of a GP with first term $ a $ and common ratio $ r $ as $ \dfrac{a\left( {{r}^{n}}-1 \right)}{r-1} $ to find $ \alpha \beta $ and then find the equation as $ {{z}^{2}}+\left( \alpha +\beta \right)z+\alpha \beta $ .


Complete step by step answer:
We are given the complex number in polar form as
\[z=\cos \left( \dfrac{2\pi }{2n+1} \right)+i\sin \left( \dfrac{2\pi }{2n+1} \right)\]
 We use the identity $ {{\left( \cos \theta +i\sin \theta \right)}^{m}}=\cos m\theta +i\sin m\theta $ where $ m $ is an integer for $ \theta =\dfrac{2\pi }{2n+1} $ and $ n=2n+1 $ to have;
\[\begin{align}
  & {{z}^{2n+1}}={{\left( \cos \left( \dfrac{2\pi }{2n+1} \right)+i\sin \left( \dfrac{2\pi }{2n+1} \right) \right)}^{2n+1}} \\
 & \Rightarrow {{z}^{2n+1}}=\cos \left( \dfrac{2\pi }{2n+1}\times 2n+1 \right)+i\sin \left( \dfrac{2\pi }{2n+1}\times 2n+1 \right) \\
 & \Rightarrow {{z}^{2n+1}}=\cos 2\pi +i\sin 2\pi =1+i\cdot 0=1 \\
\end{align}\]
We know from quadratic equations that if we are given two roots $ \alpha ,\beta $ then we can make a quadratic equation in variable $ x $ as $ {{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta =0 $ . So the required equation will be a complex quadratic equation of the form
\[{{z}^{2}}-\left( \alpha +\beta \right)z+\alpha \beta =0\]
We are given the expression for the roots $ \alpha =z+{{z}^{3}}+...+{{z}^{2n-1}} $ and $ \beta ={{z}^{2}}+{{z}^{4}}+...+{{z}^{2n}} $ . Let us find the sum of the roots as
\[\begin{align}
  & \alpha +\beta =z+{{z}^{3}}+...+{{z}^{2n-1}}+{{z}^{2}}+{{z}^{4}}+...+{{z}^{2n}} \\
 & \Rightarrow \alpha +\beta =z+{{z}^{2}}+{{z}^{3}}...+{{z}^{2n-1}}+{{z}^{2n}} \\
\end{align}\]
We add and subtract 1 in the right hand side of to have
\[\Rightarrow \alpha +\beta =1+z+{{z}^{2}}+{{z}^{3}}...+{{z}^{2n-1}}+{{z}^{2n}}-1\]
We use the sum of $ n $ terms of a GP formula for sum up to $ {{\left( 2n+1 \right)}^{th}} $ terms with first $ a=1 $ , common ratio of $ r=z $ in the right hand side to have;
\[\Rightarrow \alpha +\beta =\dfrac{1\cdot \left( {{z}^{2n+1}}-1 \right)}{z-1}-1\]
We put previously obtained $ {{z}^{2n+1}}=1 $ to have
\[\begin{align}
  & \Rightarrow \alpha +\beta =\dfrac{1\cdot \left( 1-1 \right)}{z-1}-1 \\
 & \Rightarrow \alpha +\beta =0-1=-1 \\
\end{align}\]
We express given $ \alpha =z+{{z}^{3}}+...+{{z}^{2n-1}} $ as the sum up to $ {{n}^{\text{th}}} $ term of a GP with first term $ a=z $ , common difference $ r={{z}^{2}} $ to have;
\[\alpha =\dfrac{z\left( {{z}^{n}}-1 \right)}{{{z}^{2}}-1}\]
We similarly express given $ \beta ={{z}^{2}}+{{z}^{4}}+...+{{z}^{2n}} $ as the sum up to $ {{n}^{\text{th}}} $ term of a GP with first term $ a={{z}^{2}} $ , common difference $ r={{z}^{2}} $ to have;
\[\alpha =\dfrac{{{z}^{3}}\left( {{z}^{n}}-1 \right)}{{{z}^{2}}-1}\]
So the product of the roots is
\[\begin{align}
  & \alpha \beta =\dfrac{z\left( {{z}^{n}}-1 \right)}{{{z}^{2}}-1}\cdot \dfrac{{{z}^{2}}\left( {{z}^{n}}-1 \right)}{{{z}^{2}}-1} \\
 & \Rightarrow \alpha \beta =\dfrac{{{z}^{3}}{{\left( {{z}^{n}}-1 \right)}^{2}}}{{{\left( {{z}^{2}}-1 \right)}^{2}}} \\
\end{align}\]
So the required complex equation is

\[\begin{align}
  & {{z}^{2}}-\left( -1 \right)z+\dfrac{{{z}^{3}}{{\left( {{z}^{n}}-1 \right)}^{2}}}{{{\left( {{z}^{2}}-1 \right)}^{2}}}=0 \\
 & \Rightarrow {{z}^{2}}+z+\dfrac{{{z}^{3}}{{\left( {{z}^{n}}-1 \right)}^{2}}}{{{\left( {{z}^{2}}-1 \right)}^{2}}}=0 \\
\end{align}\]

Note:
 We note that we can represent any complex number in the polar form in the complex plane as $ r{{e}^{i\theta }}=r\cos \theta +ir\sin \theta $ where $ r $ is the polar distance and $ \theta $ is the polar angle. We can alternatively find $ \alpha +\beta =-1 $ using the fact that the sum of $ {{\left( 2n+1 \right)}^{\text{th}}} $ roots of unity is zero. We note that the roots of unity are always in geometric progression.