Algebra – Complex Numbers – JEE Main 21 Jan 2026 Shift 1

Solution:

The roots of the equation $x^2 + x + 1 = 0$ are the complex cube roots of unity, $\omega$ and $\omega^2$
Let $x = \omega$. Then the general term is $\left(x^r + \frac{1}{x^r}\right)^4 = \left(\omega^r + \frac{1}{\omega^r}\right)^4 = (\omega^r + \omega^{-r})^4$.
Since $\omega^3 = 1$, we evaluate this term for different values of $r$:

Case 1: If $r$ is a multiple of 3 ($r = 3k$).
$\omega^r = 1$ and $\frac{1}{\omega^r} = 1$.
$T_r = (1 + 1)^4 = 2^4 = 16$.

Case 2: If $r$ is not a multiple of 3.
If $r = 3k+1$, $\omega^r + \frac{1}{\omega^r} = \omega + \omega^2 = -1$.
If $r = 3k+2$, $\omega^r + \frac{1}{\omega^r} = \omega^2 + \omega = -1$.
In both these cases, $T_r = (-1)^4 = 1$.

The sum runs from $r=1$ to $25$
We count how many multiples of 3 are there in $\{1, 2, \dots, 25\}$:
These are $\{3, 6, 9, 12, 15, 18, 21, 24\}$. Total = 8 terms.
The remaining terms are non-multiples of 3. Total = $25 – 8 = 17$ terms.

Total Sum = (Number of multiples of 3 $\times$ 16) + (Number of non-multiples $\times$ 1)
$$= 8(16) + 17(1)$$
$$= 128 + 17$$
$$= 145$$

Ans. (4)