Question

Let \[1,\omega ,{{\omega }^{2}}\] be the cube root of unity. The least possible degree of polynomial with real coefficients having roots \[2\omega ,\left( 2+3\omega \right),\left( 2+3{{\omega }^{2}} \right),\left( 2-\omega -{{\omega }^{2}} \right)\] is ________.

Answer 1

Hint: Apply the basic properties of cube root of unity. Thus get the conjugate of terms \[2\omega ,\left( 2+3\omega \right)\] and \[\left( 2-\omega -{{\omega }^{2}} \right)\]. Now count the complex roots and real roots, that is the least possible degree of polynomial.

Complete step-by-step solution -
The cube root of unity can be defined as the numbers which when raised to the power of 3 gives the result as 1. For example, the cube root of unity is the cube root of 1, i.e. \[\sqrt[3]{1}=1\].
Since it is given that \[1,\omega ,{{\omega }^{2}}\] are in cube root of unity, the following 3 properties hold:
\[\begin{align}
  & \left( i \right){{\omega }^{3}}=1-(1) \\
 & \left( ii \right){{\omega }^{2}}+\omega =-1-(2) \\
\end{align}\]
\[\left( iii \right)\omega ,{{\omega }^{2}}\] are conjugate to each other. – (3)
Since, \[\omega ,{{\omega }^{2}}\] are conjugate to each other then we can consider that \[a+b\omega ,a+b{{\omega }^{2}}\] are also conjugates. \[\forall a,b\in R\] and \[b\ne 0\].
Because a is only added to the real part after multiplying with b which doesn’t affect its properties.
We have been given: \[2\omega ,\left( 2+3\omega \right),\left( 2+3{{\omega }^{2}} \right),\left( 2-\omega -{{\omega }^{2}} \right)\]
We can say that as \[\left( a+b\omega \right)\] and \[\left( a+b{{\omega }^{2}} \right)\] are conjugates \[\left( 2+3\omega \right)\] and \[\left( 2+3{{\omega }^{2}} \right)\] are conjugates.
We have been given, \[2-\omega -{{\omega }^{2}}\].
From (2), \[\omega +{{\omega }^{2}}=-1\], which is one of the properties which we have mentioned before, \[1,\omega ,{{\omega }^{2}}\] are in cube root of unity.
\[\therefore {{\omega }^{2}}-\omega =1\]
Thus, \[2-\omega -{{\omega }^{2}}=2-\left( -1 \right)=2+1=3\].
Thus \[2\omega \] and \[2{{\omega }^{2}}\] are conjugates. Similarly, \[\left( 2+3\omega \right)\] and \[\left( 2+3{{\omega }^{2}} \right)\] are conjugate. Thus there are four complex roots. Now we got one real root as 3 from the above expression.
 Therefore there are 4 complex roots and 1 real root i.e. they are 5 roots. As there are 5 roots the least degree of polynomial becomes 5.
\[\therefore \] The least degree of the polynomial is 5.
Thus we got the required answer.

Note: For any polynomial equation with real coefficients, complex roots exist in pairs. Since, we already have a pair of complex roots and a real root. The conjugate of the root \[2\omega \], which is \[2{{\omega }^{2}}\] is sufficient to make all of them the roots of polynomials.