
Find the value of \[\sum\limits_{k=1}^{8}{\left[ \left( \dfrac{\cos 2k\pi }{9}+i\dfrac{\sin 2k\pi }{9} \right) \right]}\]
Answer
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Hint: Open up the summation for each value of k(i.e. from 1 to 8) and compare it with the expression for the 9th root of unity, as given below, to obtain the answer.
$1+w+{{w}^{2}}+{{w}^{3}}+{{w}^{4}}+{{w}^{5}}+{{w}^{6}}+{{w}^{7}}+{{w}^{8}}=0$
Complete step-by-step answer:
\[\sum\limits_{k=1}^{8}{\left[ \left( \dfrac{\cos 2k\pi }{9}+i\dfrac{\sin 2k\pi }{9} \right) \right]}\]
We know that 9th roots of unit are $1+w+{{w}^{2}}+{{w}^{3}}+{{w}^{4}}+{{w}^{5}}+{{w}^{6}}+{{w}^{7}}+{{w}^{8}}$
Sum of roots
$1+w+{{w}^{2}}+{{w}^{3}}+{{w}^{4}}+{{w}^{5}}+{{w}^{6}}+{{w}^{7}}+{{w}^{8}}=0$
$w+{{w}^{2}}+{{w}^{3}}+{{w}^{4}}+{{w}^{5}}+{{w}^{6}}+{{w}^{7}}+{{w}^{8}}=-1$
$\therefore \text{ if k=1}$
\[\begin{align}
& \cos \dfrac{2\left( 1 \right)\pi }{9}+i\text{ }\sin \dfrac{2\left( 1 \right)\pi }{9} \\
& =\cos \dfrac{2\pi }{9}+\text{ }i\text{ }\sin \dfrac{2\pi }{9} \\
& =w \\
\end{align}\]
If $\begin{align}
& k=2 \\
& \\
\end{align}$
\[\begin{align}
& \cos \dfrac{2\left( 2 \right)\pi }{9}+i\text{ }\sin \dfrac{2\left( 2 \right)\pi }{9} \\
& =\cos \dfrac{4\pi }{9}+\text{ }i\text{ }\sin \dfrac{4\pi }{9} \\
& ={{w}^{2}} \\
\end{align}\]
If $k=3$ then ${{w}^{3}}$
If $k=4$ then ${{w}^{4}}$
If $k=5$ then ${{w}^{5}}$
If $k=6$ then ${{w}^{6}}$
If $k=7$ then ${{w}^{7}}$
If $k=8$ then ${{w}^{8}}$
Solved as,
\[\begin{align}
& \cos \dfrac{2\left( 8 \right)\pi }{9}+i\text{ }\sin \dfrac{2\left( 8 \right)\pi }{9} \\
& =\cos \dfrac{16\pi }{9}+\text{ }i\text{ }\sin \dfrac{16\pi }{9} \\
& ={{w}^{8}} \\
\end{align}\]
The sum of all the terms
\[\sum\limits_{k=1}^{8}{\left[ \cos \left( \dfrac{2k\pi }{9} \right)+i\text{ }\sin \left( \dfrac{2k\pi }{9} \right) \right]}=-1\]
Note: $1,w\text{ and }{{\text{w}}^{2}}$ represents the cube roots of units and also
$1+w+{{w}^{2}}=0$
Some important properties of cube roots of units are:
Property 1: Among the three cube roots of unity one of the cube roots is real and the other two are conjugate complex numbers.
Property 2: Square of any one imaginary cube root of unity is equal to the other imaginary cube root of unity.
$1+w+{{w}^{2}}+{{w}^{3}}+{{w}^{4}}+{{w}^{5}}+{{w}^{6}}+{{w}^{7}}+{{w}^{8}}=0$
Complete step-by-step answer:
\[\sum\limits_{k=1}^{8}{\left[ \left( \dfrac{\cos 2k\pi }{9}+i\dfrac{\sin 2k\pi }{9} \right) \right]}\]
We know that 9th roots of unit are $1+w+{{w}^{2}}+{{w}^{3}}+{{w}^{4}}+{{w}^{5}}+{{w}^{6}}+{{w}^{7}}+{{w}^{8}}$
Sum of roots
$1+w+{{w}^{2}}+{{w}^{3}}+{{w}^{4}}+{{w}^{5}}+{{w}^{6}}+{{w}^{7}}+{{w}^{8}}=0$
$w+{{w}^{2}}+{{w}^{3}}+{{w}^{4}}+{{w}^{5}}+{{w}^{6}}+{{w}^{7}}+{{w}^{8}}=-1$
$\therefore \text{ if k=1}$
\[\begin{align}
& \cos \dfrac{2\left( 1 \right)\pi }{9}+i\text{ }\sin \dfrac{2\left( 1 \right)\pi }{9} \\
& =\cos \dfrac{2\pi }{9}+\text{ }i\text{ }\sin \dfrac{2\pi }{9} \\
& =w \\
\end{align}\]
If $\begin{align}
& k=2 \\
& \\
\end{align}$
\[\begin{align}
& \cos \dfrac{2\left( 2 \right)\pi }{9}+i\text{ }\sin \dfrac{2\left( 2 \right)\pi }{9} \\
& =\cos \dfrac{4\pi }{9}+\text{ }i\text{ }\sin \dfrac{4\pi }{9} \\
& ={{w}^{2}} \\
\end{align}\]
If $k=3$ then ${{w}^{3}}$
If $k=4$ then ${{w}^{4}}$
If $k=5$ then ${{w}^{5}}$
If $k=6$ then ${{w}^{6}}$
If $k=7$ then ${{w}^{7}}$
If $k=8$ then ${{w}^{8}}$
Solved as,
\[\begin{align}
& \cos \dfrac{2\left( 8 \right)\pi }{9}+i\text{ }\sin \dfrac{2\left( 8 \right)\pi }{9} \\
& =\cos \dfrac{16\pi }{9}+\text{ }i\text{ }\sin \dfrac{16\pi }{9} \\
& ={{w}^{8}} \\
\end{align}\]
The sum of all the terms
\[\sum\limits_{k=1}^{8}{\left[ \cos \left( \dfrac{2k\pi }{9} \right)+i\text{ }\sin \left( \dfrac{2k\pi }{9} \right) \right]}=-1\]
Note: $1,w\text{ and }{{\text{w}}^{2}}$ represents the cube roots of units and also
$1+w+{{w}^{2}}=0$
Some important properties of cube roots of units are:
Property 1: Among the three cube roots of unity one of the cube roots is real and the other two are conjugate complex numbers.
Property 2: Square of any one imaginary cube root of unity is equal to the other imaginary cube root of unity.
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