JEE Mathematics 2024 Question with Solution

Given: We need the common roots of $z^{1985} + z^{100} + 1 = 0$ and $z^2 + z + 1 = 0$.

Find: The number of values of $z$ satisfying both equations.

From $z^2 + z + 1 = 0$, the roots are the non-real cube roots of unity, namely $z = \omega$ and $z = \omega^2$, where $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.

Now reduce the exponents modulo $3$:

So for any root of $z^2 + z + 1 = 0$, we get

But from the second equation,

Hence every root of $z^2 + z + 1 = 0$ also satisfies $z^{1985} + z^{100} + 1 = 0$.

Therefore, both roots $\omega$ and $\omega^2$ are common roots. The number of common roots is $2$.

The correct option is B.

The solution contains a mismatch: the question states the second equation as $z^2 + z + 1 = 0$, while one part of the solution discusses a different equation $z^3 + 2z^2 + 2z + 1 = 0$. Using the question text and the consistent second approach in the solution gives the valid result.

If $z = \omega$, then $\omega^3 = 1$. Therefore,

So,

If $z = \omega^2$, then

Hence,

Thus both roots of $z^2 + z + 1 = 0$ are common roots. So the number of common roots is $2$. This does not match the answer key, but it is the defensible answer from the equations in the question.