JEE Mathematics 2023 Question with Solution

Given: $z_0 = \left(\frac{1}{2}, \frac{3}{2}\right)$, $z_1 = (1,1)$, and $|z_2-z_0| = \sqrt{2}$ is used in the extracted working with the points collinear.

Find: The smaller value of $|z_2|^2$.

From the solution, the line through $z_0$ and $z_1$ has

$$\tan \theta = -1$$

so

$$\theta = 135^\circ$$

Using the coordinates written in the solution,

$$z_2 = \left(\frac{1}{2} + \sqrt{2}\cos 135^\circ, \frac{3}{2} + \sqrt{2}\sin 135^\circ\right)$$

or

$$z_2 = \left(\frac{1}{2} - \sqrt{2}\cos 135^\circ, \frac{3}{2} - \sqrt{2}\sin 135^\circ\right)$$

Substituting the trigonometric values gives the two points

$$z_2 = \left(-\frac{1}{2}, \frac{5}{2}\right)$$

or

$$z_2 = \left(\frac{3}{2}, \frac{1}{2}\right)$$

Hence,

$$|z_2|^2 = \frac{26}{4}, \; \frac{5}{2}$$

So the smaller value is

$$|z_2|^2_{\min} = \frac{5}{2}$$

Therefore, the correct option is B and the smaller value of $|z_2|^2$ is $\frac{5}{2}$.

Given: The centre is $\left(\frac{1}{2}, \frac{3}{2}\right)$ and one collinear point is $z_1=(1,1)$.

Find: The smaller possible value of $|z_2|^2$.

The extracted solution treats complex numbers as points in the Argand plane. Since $z_0, z_1, z_2$ are collinear, $z_2$ lies on the same straight line through the centre. The working identifies the direction angle by

$$\tan \theta = -1$$

which gives the required direction.

Then the two possible points on that line at the stated distance from the centre are obtained by moving from the centre in opposite directions. This produces

$$\left(-\frac{1}{2}, \frac{5}{2}\right)$$

and

$$\left(\frac{3}{2}, \frac{1}{2}\right)$$

Now compute modulus squared as $x^2+y^2$ for each point:

$$\left(-\frac{1}{2}\right)^2 + \left(\frac{5}{2}\right)^2 = \frac{1}{4}+\frac{25}{4}=\frac{26}{4}=\frac{13}{2}$$ $$\left(\frac{3}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{9}{4}+\frac{1}{4}=\frac{10}{4}=\frac{5}{2}$$

Among these, the smaller value is $\frac{5}{2}$. Hence the correct option is B.