JEE Mathematics 2025 Question with Solution

Given: $A = \{ z \in \mathbb{C} : |z - 2 - i| = 3 \}$, $B = \{ z \in \mathbb{C} : \operatorname{Re}(z - iz) = 2 \}$, and $S = A \cap B$.

Find: $\sum_{z \in S} |z|^2$.

Let $z = x + iy$.

From

$$|z - 2 - i| = 3$$

we get

$$|(x-2) + (y-1)i| = 3$$

so

$$(x - 2)^2 + (y - 1)^2 = 9.$$

Also,

$$\operatorname{Re}(z - iz) = 2.$$

Since

$$z - iz = (1-i)(x+iy) = (x+y) + i(y-x),$$

its real part is $x+y$. Hence

$$x + y = 2.$$

Now substitute $y = 2-x$ into the circle equation:

$$(x - 2)^2 + (2 - x - 1)^2 = 9$$ $$(x - 2)^2 + (1 - x)^2 = 9$$ $$x^2 - 4x + 4 + 1 - 2x + x^2 = 9$$ $$2x^2 - 6x + 5 = 9$$ $$2x^2 - 6x - 4 = 0$$ $$x^2 - 3x - 2 = 0.$$

Therefore,

$$x = \frac{3 \pm \sqrt{17}}{2}.$$

Then the corresponding values of $y$ are

$$y = 2 - x = \frac{1 \mp \sqrt{17}}{2}.$$

So the two points of intersection are obtained, and

$$\sum_{z \in S} |z|^2 = \left( \frac{3 + \sqrt{17}}{2} \right)^2 + \left( \frac{1 - \sqrt{17}}{2} \right)^2 + \left( \frac{3 - \sqrt{17}}{2} \right)^2 + \left( \frac{1 + \sqrt{17}}{2} \right)^2.$$

Thus,

$$\sum_{z \in S} |z|^2 = \frac{1}{4}\left[2 \times 26 + 2 \times 18\right] = \frac{88}{4} = 22.$$

Therefore, the correct option is A.

Given: The intersection points satisfy

$$(x - 2)^2 + (y - 1)^2 = 9$$

and

$$x + y = 2.$$

Find: $\sum_{z \in S} |z|^2$.

Using $y = 2-x$ in the circle equation gives

$$x^2 - 3x - 2 = 0.$$

Let its roots be $x_1, x_2$. Then

$$x_1 + x_2 = 3, \qquad x_1x_2 = -2.$$

Also, for each intersection point, $y_i = 2 - x_i$.

Now

$$|z_i|^2 = x_i^2 + y_i^2 = (x_i + y_i)^2 - 2x_i y_i = 4 - 2x_i y_i.$$

Hence

$$\sum |z_i|^2 = 8 - 2\sum x_i y_i.$$

But

$$x_i y_i = x_i(2-x_i) = 2x_i - x_i^2,$$

so

$$\sum x_i y_i = 2(x_1+x_2) - (x_1^2 + x_2^2).$$

Now

$$x_1^2 + x_2^2 = (x_1+x_2)^2 - 2x_1x_2 = 3^2 - 2(-2) = 13.$$

Therefore,

$$\sum x_i y_i = 2\cdot 3 - 13 = -7.$$

So,

$$\sum |z|^2 = 8 - 2(-7) = 22.$$

Therefore, the correct option is A.