JEE Mathematics 2025 Question with Solution

Given: Integers $a, b \in [-3,3]$ with $a+b \neq 0$.

We need the number of ordered pairs $(a,b)$ for which

and the determinant condition holds.

Find: The total number of valid ordered pairs.

From

we get

This means $z$ is equidistant from the real points $a$ and $-b$, so the locus is the perpendicular bisector of the segment joining them. Hence

Using the determinant condition and the symmetry involving cube roots of unity, the solution simplifies it to the circle

So the locus of $z$ is the circle with center $\,(1,0)\,$ and radius $1$.

For the vertical line

to intersect the circle

its distance from the center $\,(1,0)\,$ must not exceed the radius:

Therefore,

which gives

and hence

Now count integer pairs $(a,b)$ with $a,b \in [-3,3]$ satisfying

and also

From the extracted counting table in the solution, the valid counts are:

• for $a=0$: $3$ pairs,
• for $a=1$: $4$ pairs,
• for $a=2$: $4$ pairs,
• for $a=3$: $4$ pairs, with the final overlap adjustment shown there.

Thus, as concluded in the provided solution,

Therefore, the required number of ordered pairs is $10$.

Given: $\left|\frac{z-a}{z+b}\right|=1$ and the determinant condition.

Find: How these conditions restrict $a$ and $b$.

The first condition becomes

Geometrically, this says the point $z$ lies on the perpendicular bisector of the points representing $a$ and $-b$ on the real axis. Therefore the real part of $z$ is fixed:

The second condition is stated in the extracted solution to simplify to

So $z$ must also lie on the circle centered at $1$ with radius $1$.

Hence we need the line

to meet the circle

That is possible exactly when

which reduces to

Together with $a,b\in[-3,3]$ and $a+b\ne 0$, the extracted solution counts the admissible ordered pairs and obtains $10$.