Question

If ${{x}^{2}}-x+1=0$, then find the value of $\sum\limits_{n=1}^{5}{{{\left( {{x}^{n}}+\dfrac{1}{{{x}^{n}}} \right)}^{2}}}$.

Answer 1

Hint: In this problem, we will first need to find the roots of the quadratic equation using the quadratic formula. Then, we can invert the value of $x$ to find $\dfrac{1}{x}$, and find $\theta $ by changing them into polar form. Now, we should use the DeMoivre’s theorem that states ${{\left[ r\left( \cos \phi +i\sin \phi \right) \right]}^{n}}={{r}^{n}}\left( \cos n\phi +i\sin n\phi \right)$, we can then further expand the summation and substitute the value of $\theta $ to get the answer.

Complete step by step answer:
We are given with the quadratic equation ${{x}^{2}}-x+1=0$. Let us find the roots of this equation.
We know by the quadratic formula that the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ are $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
So, here we will have
\[x=\dfrac{1\pm \sqrt{1-4}}{2}\]
\[\Rightarrow x=\dfrac{1\pm \sqrt{3}i}{2}\]
So, we can also write
\[\dfrac{1}{x}=\dfrac{2}{1\pm \sqrt{3}i}\]
We can simplify the above equation by rationalising as
\[\dfrac{1}{x}=\dfrac{2}{1\pm \sqrt{3}i}\times \dfrac{1\mp \sqrt{3}i}{1\mp \sqrt{3}i}\]
\[\Rightarrow \dfrac{1}{x}=\dfrac{2\left( 1\mp \sqrt{3}i \right)}{{{\left( 1 \right)}^{2}}-{{\left( \sqrt{3}i \right)}^{2}}}\]
We can further simplify this as,
\[\dfrac{1}{x}=\dfrac{2\left( 1\mp \sqrt{3}i \right)}{1-\left( -3 \right)}\]
\[\Rightarrow \dfrac{1}{x}=\dfrac{1\mp \sqrt{3}i}{2}\]
Here, we can see that $x\text{ and }\dfrac{1}{x}$ are complex numbers. We also know that we can represent complex numbers as $z=r\cos \theta +ir\sin \theta $, where r is the magnitude.
For $x\text{ and }\dfrac{1}{x}$, the magnitudes are = $\sqrt{{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}}=\sqrt{\dfrac{1}{4}+\dfrac{3}{4}}=1$
Since, magnitude is 1, we can safely assume \[x=\cos \theta +i\sin \theta \], and so we can write
$\cos \theta +i\sin \theta =\dfrac{1}{2}\pm i\dfrac{\sqrt{3}}{2}$
Thus, we now have $\cos \theta =\dfrac{1}{2}\text{ and }\sin \theta =\pm \dfrac{\sqrt{3}}{2}$.
So, here we can conclude that $\theta =\pm \dfrac{\pi }{3}$ .
So, \[x=\cos \dfrac{\pi }{3}\pm i\sin \dfrac{\pi }{3}\text{ and }\dfrac{1}{x}=\cos \dfrac{\pi }{3}\mp i\sin \dfrac{\pi }{3}\].
We know that according to DeMoivre’s theorem, if $z=\cos \phi +i\sin \phi $ , then
${{z}^{n}}=\cos n\phi +i\sin n\phi $.
Using DeMoivre’s theorem, we get
$\begin{align}
  & {{x}^{n}}=\cos \dfrac{n\pi }{3}\pm i\sin \dfrac{n\pi }{3} \\
 & \dfrac{1}{{{x}^{n}}}=\cos \dfrac{n\pi }{3}\mp i\sin \dfrac{n\pi }{3} \\
\end{align}$
Adding these two equations, we get
${{x}^{n}}+\dfrac{1}{{{x}^{n}}}=2\cos \dfrac{n\pi }{3}$
On squaring both sides of the equation, we get
\[{{\left( {{x}^{n}}+\dfrac{1}{{{x}^{n}}} \right)}^{2}}={{\left( 2\cos \dfrac{n\pi }{3} \right)}^{2}}\]
\[\Rightarrow {{\left( {{x}^{n}}+\dfrac{1}{{{x}^{n}}} \right)}^{2}}=4{{\cos }^{2}}\dfrac{n\pi }{3}\]
So, now we can easily write the equation
$\sum\limits_{n=1}^{5}{{{\left( {{x}^{n}}+\dfrac{1}{{{x}^{n}}} \right)}^{2}}}=\sum\limits_{n=1}^{5}{\left( 4{{\cos }^{2}}\dfrac{n\pi }{3} \right)}$
The term 4 in the RHS part can be brought out of the summation,
$\sum\limits_{n=1}^{5}{{{\left( {{x}^{n}}+\dfrac{1}{{{x}^{n}}} \right)}^{2}}}=4\sum\limits_{n=1}^{5}{\left( {{\cos }^{2}}\dfrac{n\pi }{3} \right)}$
To calculate this summation on the RHS part, we can expand it by putting different values of n,
$\sum\limits_{n=1}^{5}{{{\left( {{x}^{n}}+\dfrac{1}{{{x}^{n}}} \right)}^{2}}}=4\left( {{\cos }^{2}}\dfrac{\pi }{3}+{{\cos }^{2}}\dfrac{2\pi }{3}+{{\cos }^{2}}\dfrac{3\pi }{3}+{{\cos }^{2}}\dfrac{4\pi }{3}+{{\cos }^{2}}\dfrac{5\pi }{3} \right)$
Now, from trigonometry, we know that
$\begin{align}
  & \cos \dfrac{\pi }{3}=\dfrac{1}{2} \\
 & \cos \dfrac{2\pi }{3}=-\dfrac{1}{2} \\
 & \cos \dfrac{3\pi }{3}=-1 \\
 & \cos \dfrac{4\pi }{3}=-\dfrac{1}{2} \\
 & \cos \dfrac{5\pi }{3}=\dfrac{1}{2} \\
\end{align}$
Squaring each of these values, we get
$\begin{align}
  & {{\cos }^{2}}\dfrac{\pi }{3}=\dfrac{1}{4} \\
 & {{\cos }^{2}}\dfrac{2\pi }{3}=\dfrac{1}{4} \\
 & {{\cos }^{2}}\dfrac{3\pi }{3}=1 \\
 & {{\cos }^{2}}\dfrac{4\pi }{3}=\dfrac{1}{4} \\
 & {{\cos }^{2}}\dfrac{5\pi }{3}=\dfrac{1}{4} \\
\end{align}$
Using these values in our original equation, we get
$\sum\limits_{n=1}^{5}{{{\left( {{x}^{n}}+\dfrac{1}{{{x}^{n}}} \right)}^{2}}}=4\left( \dfrac{1}{4}+\dfrac{1}{4}+1+\dfrac{1}{4}+\dfrac{1}{4} \right)$
$\Rightarrow \sum\limits_{n=1}^{5}{{{\left( {{x}^{n}}+\dfrac{1}{{{x}^{n}}} \right)}^{2}}}=4\times 2$
$\Rightarrow \sum\limits_{n=1}^{5}{{{\left( {{x}^{n}}+\dfrac{1}{{{x}^{n}}} \right)}^{2}}}=8$

Thus, the value of $\sum\limits_{n=1}^{5}{{{\left( {{x}^{n}}+\dfrac{1}{{{x}^{n}}} \right)}^{2}}}$ is 8.

Note: We must remember that if magnitude of complex number is not 1, then, according to DeMoivre’s theorem, ${{\left[ r\left( \cos \phi +i\sin \phi \right) \right]}^{n}}={{r}^{n}}\left( \cos n\phi +i\sin n\phi \right)$. We must also note the difference between the two signs $\pm \text{ and }\mp $.