Question

If \[\omega \] is a non real cube root of unity and n is a positive integer which is not a multiple of 3, then $1+{{\omega }^{n}}+{{\omega }^{2n}}$ is equal to:
A. $3\omega $
B. 0
C. 3
D. None of these

Answer 1

Hint: Using b=3 in Euclid’s division lemma, take possible cases for $n=3k+r$ and for each case solve the given expression using properties of exponents. After solving you will get $1+{{\omega }^{n}}+{{\omega }^{2n}}=1+\omega +{{\omega }^{2}}$ .Use the formula for sum of the roots in the equation ${{x}^{3}}-1=0$ to get the value of $1+\omega +{{\omega }^{2}}$.

Complete step-by-step answer:
Given:
‘\[\omega \]’ is a non real cube root of unity.
i.e. \[{{\omega }^{3}}=1\] ……………………(1)
And we have to find the value of $1+{{\omega }^{n}}+{{\omega }^{2n}}$ . is also given that ‘n’ is not a multiple of 3. By Euclid’s division lemma, we can write that $a=bk+r; 0\le {r}<{b}$ . for any positive integer ‘b’ .
Let us take b=3.
So, $a=3k+r;0\le{r}<3$ .
Three possible cases:-
Case (1): $a=3k+0$
Case (2): $a=3k+1$
Case (3): $a=3k+2$ .
In case ${{1}^{st}}$ ‘a’ is a multiple of 3 but here, it is given that ‘n’ is not a multiple of 3. For ‘n’, there are two possible cases.
Case (1): $n=3k+1$ and
Case (2): $n=3k+2$ .
Let us first solve the given expression for case (1) i.e. $n=3k+1$
$\begin{align}
  & 1+{{\omega }^{n}}+{{\omega }^{2n}}=1+{{\omega }^{\left( 3k+1 \right)}}+{{\omega }^{2\left( 3k+1 \right)}} \\
 & \Rightarrow 1+{{\omega }^{n}}+{{\omega }^{2n}}=1+{{\omega }^{\left( 3k+1 \right)}}+{{\omega }^{\left( 6k+2 \right)}} \\
\end{align}$
We know for any integers ${{a}^{b+c}}={{a}^{b}}\times {{a}^{c}}$ .
So, we will get
$\Rightarrow 1+{{\omega }^{n}}+{{\omega }^{2n}}=1+{{\omega }^{3k}}.{{\omega }^{1}}+{{\omega }^{6k}}.{{\omega }^{2}}$
We know for any integers ${{a}^{bc}}={{\left( {{a}^{b}} \right)}^{c}}$ .
So, we will get
$\Rightarrow 1+{{\omega }^{n}}+{{\omega }^{2n}}=1+{{\left( {{\omega }^{3}} \right)}^{k}}.{{\omega }^{1}}+{{\left( {{\omega }^{3}} \right)}^{2k}}.{{\omega }^{2}}$
From eq (1) , \[{{\omega }^{3}}=1\] .
On putting \[{{\omega }^{3}}=1\], we will get,
$\Rightarrow 1+{{\omega }^{n}}+{{\omega }^{2n}}=1+{{\left( 1 \right)}^{k}}.\omega +{{\left( 1 \right)}^{2k}}.{{\omega }^{2}}$ .
We know for any integer a, ${{1}^{a}}=1$ .
So, we will get,
$\Rightarrow 1+{{\omega }^{n}}+{{\omega }^{2n}}=1+\omega +{{\omega }^{2}}$ .
Now, let us similarly solve the given expression for case (2) i.e. $b=3k+2$ .
$1+{{\omega }^{n}}+{{\omega }^{2n}}=1+{{\omega }^{\left( 3k+2 \right)}}+{{\omega }^{2\left( 3k+2 \right)}}$ .
Using the properties that we have used in solving for case (1), we get
$\Rightarrow 1+{{\omega }^{n}}+{{\omega }^{2n}}=1+{{\omega }^{3}}^{k}.{{\omega }^{2}}+{{\omega }^{\left( 6k+4 \right)}}$ .
We can write $6k+4=6k+3+1$ .
So, we will get
\[\begin{align}
  & \Rightarrow 1+{{\omega }^{n}}+{{\omega }^{2n}}=1+{{\left( {{\omega }^{3}} \right)}^{k}}.{{\omega }^{2}}+{{\omega }^{\left( 6k+3+1 \right)}} \\
 & \Rightarrow 1+{{\omega }^{n}}+{{\omega }^{2n}}=1+{{\left( {{\omega }^{3}} \right)}^{k}}.{{\omega }^{2}}+{{\omega }^{1}}.{{\omega }^{3+6k}} \\
 & \Rightarrow 1+{{\omega }^{n}}+{{\omega }^{2n}}=1+{{\left( {{\omega }^{3}} \right)}^{k}}.{{\omega }^{2}}+{{\omega }^{1}}.{{\left( {{\omega }^{3}} \right)}^{\left( 1+2k \right)}} \\
\end{align}\]
From eq (1) \[{{\omega }^{3}}=1\] .
On putting \[{{\omega }^{3}}=1\], we will get,
$\begin{align}
  & \Rightarrow 1+{{\omega }^{n}}+{{\omega }^{2n}}=1+{{\left( 1 \right)}^{k}}.{{\omega }^{2}}+{{\omega }^{1}}.{{\left( 1 \right)}^{1+2k}} \\
 & \Rightarrow 1+{{\omega }^{n}}+{{\omega }^{2n}}=1+{{\omega }^{2}}+\omega \\
 & \Rightarrow 1+{{\omega }^{n}}+{{\omega }^{2n}}=1+\omega +{{\omega }^{2}} \\
\end{align}$
In both the cases, we have obtained that
$\Rightarrow 1+{{\omega }^{n}}+{{\omega }^{2n}}=1+\omega +{{\omega }^{2}}$…………………………. (2)
We know that $1,\omega ,{{\omega }^{2}}$ are the three roots of the equation ${{x}^{3}}+1$ .
i.e. ${{x}^{3}}-1=0$ .
For a cubic equation $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$ .
We know that sum of roots \[=\dfrac{\text{-}\left( \text{coefficient of }{{\text{x}}^{2}} \right)}{\text{coefficient of }{{\text{x}}^{3}}}\] .
Thus, $1+\omega +{{\omega }^{2}}=\dfrac{\text{-}\left( \text{coefficient of }{{\text{x}}^{2}} \right)}{\text{coefficient of }{{\text{x}}^{3}}}$
Here in equation ${{x}^{3}}-1=0$ , coefficient of ${{x}^{2}}=0$ and coefficient of ${{x}^{3}}=1$ .
So, $1+\omega +{{\omega }^{2}}=\dfrac{\text{-}\left( 0 \right)}{1}$ .
$\Rightarrow 1+\omega +{{\omega }^{2}}=0$ ………………. (3)
So, $1+{{\omega }^{n}}+{{\omega }^{2n}}=0$ [from eq (2) and (3)].
Hence the required value of $1+{{\omega }^{n}}+{{\omega }^{2n}}=0$ and option (B) is the correct answer.

Note: (i) According to Euclid's Division Lemma if we have two positive integers a and b, then there exist unique integers q and r which satisfies the condition $a = bq + r$ where $0\le {r}< {b}$. ... By exactly we mean that on dividing both the integers a and b the remainder is zero
(ii) Shortcut method:
 If ‘\[\omega \]’ is a non real root of unity then,
$1+{{\omega }^{n}}+{{\omega }^{2n}}=0$ if ‘n’ is not a multiple of 3.
$1+{{\omega }^{n}}+{{\omega }^{2n}}=3$ if ‘n’ is a multiple of 3.
In the solution, we have proved that if ‘n’ is not a multiple of 3, $1+{{\omega }^{n}}+{{\omega }^{2n}}=0$.
Now let us prove if ‘b’ is a multiple of 3, $1+{{\omega }^{n}}+{{\omega }^{2n}}=3$ .
Since ‘b’ is a multiple of 3, $n=3k$ .
Putting $n=3k$ , we will get,
$1+{{\omega }^{n}}+{{\omega }^{2n}}=1+{{\omega }^{3k}}+{{\omega }^{2\left( 3k \right)}}$
$\begin{align}
  & =1+{{\omega }^{3k}}+{{\omega }^{6k}} \\
 & =1+{{\left( {{\omega }^{3}} \right)}^{k}}+{{\left( {{\omega }^{3}} \right)}^{2k}} \\
\end{align}$
One putting ${{\omega }^{3}}=1$ , we will get,
$\begin{align}
  & 1+{{\omega }^{n}}+{{\omega }^{2n}}=1+{{1}^{k}}+{{\left( 1 \right)}^{2k}} \\
 & \Rightarrow 1+{{\omega }^{n}}+{{\omega }^{2n}}=1+1+1 \\
\end{align}$
$\Rightarrow 1+{{\omega }^{n}}+{{\omega }^{2n}}=3$.