Question

If \[\omega \] is an imaginary cube root of unity then the value of \[\cos \left[ \dfrac{\pi }{225}\sum\limits_{r=1}^{10}{\left( r-\omega \right)\left( r-\overline{\omega } \right)} \right]\] is: -
(a) -1
(b) \[\dfrac{-1}{2}\]
(c) \[\dfrac{1}{2}\]
(d) 1
(e) 0

Answer 1

Hint: First of all use the conversion: - \[\overline{\omega }={{\omega }^{2}}\]. Now, multiply the terms \[\left( r-\omega \right)\left( r-\overline{\omega } \right)\] and use the formulas: - \[1+\omega +{{\omega }^{2}}=0\] and \[{{\omega }^{3}}=1\] to simplify the product. Take the summation of these obtained products and use the formulas: -\[\sum\limits_{r=1}^{n}{{{r}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6},\sum\limits_{r=1}^{n}{r}=\dfrac{n\left( n+1 \right)}{2}\] and \[\sum\limits_{r=1}^{n}{k}=nk\], where k = constant, to get the required sum. Multiply this sum with \[\dfrac{\pi }{225}\] and take cosine function to get the answer.

Complete step by step answer:
Here, we have been provided with an imaginary cube root of unity \[\omega \] and we have been asked to find the value of the expression: - \[\cos \left[ \dfrac{\pi }{225}\sum\limits_{r=1}^{10}{\left( r-\omega \right)\left( r-\overline{\omega } \right)} \right]\].
Now, we know that \[\omega =\dfrac{-1+\sqrt{3}i}{2}\] and its conjugate (\[\overline{\omega }\]) is given as: -
\[\Rightarrow \overline{\omega }=\dfrac{-1-\sqrt{3}i}{2}={{\omega }^{2}}\]
\[\Rightarrow \overline{\omega }={{\omega }^{2}}\] - (1)
Now, let us find the value of the expression: - \[\sum\limits_{r=1}^{10}{\left( r-\omega \right)\left( r-\overline{\omega } \right)}\]. So, using equation (1), we get,
\[\Rightarrow \sum\limits_{r=1}^{10}{\left( r-\omega \right)\left( r-\overline{\omega } \right)}=\sum\limits_{r=1}^{10}{\left( r-\omega \right)\left( r-{{\omega }^{2}} \right)}\]
\[\begin{align}
  & \Rightarrow \sum\limits_{r=1}^{10}{\left( r-\omega \right)\left( r-\overline{\omega } \right)}=\sum\limits_{r=1}^{10}{\left( {{r}^{2}}-r\omega -r{{\omega }^{2}}+{{\omega }^{3}} \right)} \\
 & \Rightarrow \sum\limits_{r=1}^{10}{\left( r-\omega \right)\left( r-\overline{\omega } \right)}=\sum\limits_{r=1}^{10}{\left( {{r}^{2}}-r\left( \omega +{{\omega }^{2}} \right)+{{\omega }^{3}} \right)} \\
\end{align}\]
Now, applying the formulas: - \[1+\omega +{{\omega }^{2}}=0\] and \[{{\omega }^{3}}=1\], we get,
\[\begin{align}
  & \Rightarrow \sum\limits_{r=1}^{10}{\left( r-\omega \right)\left( r-\overline{\omega } \right)}=\sum\limits_{r=1}^{10}{\left( {{r}^{2}}-r\left( -1 \right)+1 \right)} \\
 & \Rightarrow \sum\limits_{r=1}^{10}{\left( r-\omega \right)\left( r-\overline{\omega } \right)}=\sum\limits_{r=1}^{10}{\left( {{r}^{2}}+r+1 \right)} \\
\end{align}\]
Breaking the terms with summation sign, we get,
\[\Rightarrow \sum\limits_{r=1}^{10}{\left( r-\omega \right)\left( r-\overline{\omega } \right)}=\sum\limits_{r=1}^{10}{{{r}^{2}}}+\sum\limits_{r=1}^{10}{r}+\sum\limits_{r=1}^{10}{1}\]
Here, we can clearly, see that the first term in R.H.S., i.e. \[\sum\limits_{r=1}^{10}{{{r}^{2}}}\] is the summation of squares of first 10 natural numbers whose sum is given as \[\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\], where n = 10. The second term is the summation of first 10 natural numbers, i.e. \[\sum\limits_{r=1}^{10}{r}\], whose sum is given as \[\dfrac{n\left( n+1 \right)}{2}\], where n = 10. Now, the third term is the summation of a constant k = 1, i.e. \[\sum\limits_{r=1}^{10}{1}\], whose sum is given as n, where n = 10. So, simplifying the summation, we get,
\[\begin{align}
  & \Rightarrow \sum\limits_{r=1}^{10}{\left( r-\omega \right)\left( r-\overline{\omega } \right)}=\dfrac{10\times \left( 10+1 \right)\times \left( 20+1 \right)}{6}+\dfrac{10\times \left( 10+1 \right)}{2}+10 \\
 & \Rightarrow \sum\limits_{r=1}^{10}{\left( r-\omega \right)\left( r-\overline{\omega } \right)}=450 \\
\end{align}\]
Multiplying \[\dfrac{\pi }{225}\] both sides, we get,
\[\begin{align}
  & \Rightarrow \dfrac{\pi }{225}\sum\limits_{r=1}^{10}{\left( r-\omega \right)\left( r-\overline{\omega } \right)}=\dfrac{\pi }{225}\times 450 \\
 & \Rightarrow \dfrac{\pi }{225}\sum\limits_{r=1}^{10}{\left( r-\omega \right)\left( r-\overline{\omega } \right)}=2\pi \\
\end{align}\]
Now, taking cosine function both sides, we get,
\[\Rightarrow \cos \left[ \dfrac{\pi }{225}\sum\limits_{r=1}^{10}{\left( r-\omega \right)\left( r-\overline{\omega } \right)} \right]=\cos 2\pi \]
\[\Rightarrow \cos \left[ \dfrac{\pi }{225}\sum\limits_{r=1}^{10}{\left( r-\omega \right)\left( r-\overline{\omega } \right)} \right]=1\]
Hence, option (d) is our answer.

Note:
 One must know the meaning of conjugate of a complex number to solve the question because without using the equality \[\overline{\omega }={{\omega }^{2}}\] we cannot use the formulas: - \[1+\omega +{{\omega }^{2}}=0\] and \[{{\omega }^{3}}=1\] to simplify the summation. You must not try to substitute the value of \[\omega \] and \[{{\omega }^{2}}\] and then evaluate the summation as it will make our calculation difficult.