
The minimum value of \[\left| a+b\omega +c{{\omega }^{2}} \right|\] where \[a,b,c\] are all not equal integers and \[\omega \left( \ne 1 \right)\] is a cube root of unity, is
1) \[\sqrt{3}\]
2) \[\dfrac{1}{2}\]
3) \[1\]
4) \[0\]
Answer
510.3k+ views
Hint: In this type of question we have to use the concept of the cube root of unity. We know that the cube root of unity is represented by \[\omega \]. Also we know that there are three cube roots of unity namely \[1,\omega ,{{\omega }^{2}}\] and their sum is equal to zero i.e. \[1+\omega +{{\omega }^{2}}=0\].
Complete step-by-step solution:
Now we have to find the value of \[\left| a+b\omega +c{{\omega }^{2}} \right|\] where \[a,b,c\] are all not equal integers and \[\omega \left( \ne 1 \right)\] is a cube root of unity
For this let us consider
\[\Rightarrow z=\left| a+b\omega +c{{\omega }^{2}} \right|\]
Now as we know that there are three cube roots of unity namely \[1,\omega ,{{\omega }^{2}}\] and their sum is equal to zero i.e. \[1+\omega +{{\omega }^{2}}=0\]
\[\Rightarrow {{\omega }^{2}}=-1-\omega \]
By substituting this in above expression we get,
\[\begin{align}
& \Rightarrow z=\left| a+b\omega -c-c\omega \right| \\
& \Rightarrow z=\left| \left( a-c \right)+\left( b-c \right)\omega \right| \\
\end{align}\]
\[\Rightarrow z\ge \left| a-c \right|+\left| \left( b-c \right)\omega \right|\]
Now if \[\omega =1\] then we get
\[\Rightarrow z=\left| a-c \right|+\left| \left( b-c \right) \right|\]
Now for \[z\] to be minimum
\[\Rightarrow \left| a-c \right|=0,\left| b-c \right|=0\]
\[\Rightarrow a=c,b=c\]
Which is not possible as we have given that \[a,b,c\] are all not equal integers
Thus for \[z\] to be minimum we have
\[\Rightarrow \text{If }\left| a-c \right|=0\text{ then }\left| b-c \right|=1\]
Hence, we can write
\[\begin{align}
& \Rightarrow z\ge \left| a-c \right|+\left| \left( b-c \right)\omega \right| \\
& \Rightarrow z\ge 1 \\
\end{align}\]
Thus, option (3) is the correct option.
Note:In this question students must be familiar with the cube root of unity. Students have to note that to represent the cube root of unity we use the symbol \[\omega \]. The cube root of unity has three values out of which one value is 1 and the remaining two are the complex values which we denote as \[\omega \] and \[{{\omega }^{2}}\].
Complete step-by-step solution:
Now we have to find the value of \[\left| a+b\omega +c{{\omega }^{2}} \right|\] where \[a,b,c\] are all not equal integers and \[\omega \left( \ne 1 \right)\] is a cube root of unity
For this let us consider
\[\Rightarrow z=\left| a+b\omega +c{{\omega }^{2}} \right|\]
Now as we know that there are three cube roots of unity namely \[1,\omega ,{{\omega }^{2}}\] and their sum is equal to zero i.e. \[1+\omega +{{\omega }^{2}}=0\]
\[\Rightarrow {{\omega }^{2}}=-1-\omega \]
By substituting this in above expression we get,
\[\begin{align}
& \Rightarrow z=\left| a+b\omega -c-c\omega \right| \\
& \Rightarrow z=\left| \left( a-c \right)+\left( b-c \right)\omega \right| \\
\end{align}\]
\[\Rightarrow z\ge \left| a-c \right|+\left| \left( b-c \right)\omega \right|\]
Now if \[\omega =1\] then we get
\[\Rightarrow z=\left| a-c \right|+\left| \left( b-c \right) \right|\]
Now for \[z\] to be minimum
\[\Rightarrow \left| a-c \right|=0,\left| b-c \right|=0\]
\[\Rightarrow a=c,b=c\]
Which is not possible as we have given that \[a,b,c\] are all not equal integers
Thus for \[z\] to be minimum we have
\[\Rightarrow \text{If }\left| a-c \right|=0\text{ then }\left| b-c \right|=1\]
Hence, we can write
\[\begin{align}
& \Rightarrow z\ge \left| a-c \right|+\left| \left( b-c \right)\omega \right| \\
& \Rightarrow z\ge 1 \\
\end{align}\]
Thus, option (3) is the correct option.
Note:In this question students must be familiar with the cube root of unity. Students have to note that to represent the cube root of unity we use the symbol \[\omega \]. The cube root of unity has three values out of which one value is 1 and the remaining two are the complex values which we denote as \[\omega \] and \[{{\omega }^{2}}\].
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