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:

& \left( i \right){{\omega }^{3}}=1-(1) \\\ & \left( ii \right){{\omega }^{2}}+\omega =-1-(2) \\\ \end{aligned}$$ $$\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.