Complex Numbers – Cube Roots of Unity – JEE Main 22 Jan 2026 Shift 1

Solution:

We recognize that $\alpha = \omega$ and $\beta = \omega^2$, where $\omega$ is the complex cube root of unity.
Properties of $\omega$: $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.

Let the four terms in the expression be $T_1^{20}, T_2^{20}, T_3^{20}, T_4^{20}$.
$$ T_1 = 7 – 7\omega + 9\omega^2 $$
$$ T_2 = 9 + 7\omega – 7\omega^2 $$
$$ T_3 = -7 + 9\omega + 7\omega^2 $$
$$ T_4 = 14 + 7\alpha + 7\beta = 14 + 7(\omega + \omega^2) $$

Simplifying $T_4$:
Since $\omega + \omega^2 = -1$,
$$ T_4 = 14 + 7(-1) = 7 $$

Now let’s establish a relationship between $T_1, T_2, T_3$.
Consider $\omega T_2$:
$$ \omega T_2 = \omega(9 + 7\omega – 7\omega^2) = 9\omega + 7\omega^2 – 7\omega^3 = 9\omega + 7\omega^2 – 7 $$
This is exactly $T_3$. So, $T_3 = \omega T_2$.

Consider $\omega^2 T_2$:
$$ \omega^2 T_2 = \omega^2(9 + 7\omega – 7\omega^2) = 9\omega^2 + 7\omega^3 – 7\omega^4 = 9\omega^2 + 7 – 7\omega $$
This is exactly $T_1$. So, $T_1 = \omega^2 T_2$.

Now substitute these back into the main expression:
$$ S = (\omega^2 T_2)^{20} + (T_2)^{20} + (\omega T_2)^{20} + (7)^{20} $$
$$ S = T_2^{20}(\omega^{40} + 1 + \omega^{20}) + 7^{20} $$

Using $\omega^3 = 1$:
$$ S = T_2^{20}(1 + \omega + \omega^2) + 7^{20} $$
Since $1 + \omega + \omega^2 = 0$, the first part becomes 0.
$$ S = 0 + 7^{20} = 7^{20} $$

We are given $S = m^{10}$.
$$ m^{10} = 7^{20} = (7^2)^{10} = 49^{10} $$
$$ m = 49 $$

Ans. (49)