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JEE Mathematics 2024 Question with Solution

Let SS = {z∈Cz \in C : ∣z−1∣=1|z - 1| = 1 and (2−1)(z+zˉ)=(zˉ−z)=22(\sqrt{2} - 1)(z + \bar{z}) = (\bar{z} - z) = 2\sqrt{2}}. Let z1,z2∈Sz_1, z_2 \in S be such that ∣z1∣=max⁡z∈S∣z∣|z_1| = \max_{z \in S} |z| and ∣z2∣=min⁡z∈S∣z∣|z_2| = \min_{z \in S} |z|. Then ∣z1−z2∣2\sqrt{|z_1 - z_2|^2} equals:

  • A

    11

  • B

    44

  • C

    33

  • D

    22

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: S={z∈C:∣z−1∣=1}S = \{z \in C : |z-1|=1\} together with the linear condition used in the solution.

Find: ∣z1−z2∣2\sqrt{|z_1-z_2|^2} where ∣z1∣|z_1| is maximum and ∣z2∣|z_2| is minimum over SS.

Let

Z=x+iyZ = x + iy

Then

(x−1)2+y2=1⋯(1)(x-1)^2 + y^2 = 1 \qquad \cdots (1)

and

(2−1)(2x)+i(2y)=22(\sqrt{2}-1)(2x) + i(2y) = 2\sqrt{2}

which gives

(2−1)x+y=2⋯(2)(\sqrt{2}-1)x + y = \sqrt{2} \qquad \cdots (2)

Solving equations (1) and (2), we get

x=1orx=−12⋯(3)x = 1 \quad \text{or} \quad x = -\frac{1}{\sqrt{2}} \qquad \cdots (3)

Now solving (3) with (2):

For x=1,  y=1⇒Z1=1+i\text{For } x=1, \; y=1 \Rightarrow Z_1 = 1+i

and

For x=−12,  y=2−12⇒Z2=−12+i(2−12)\text{For } x=-\frac{1}{\sqrt{2}}, \; y = \sqrt{2}-\frac{1}{\sqrt{2}} \Rightarrow Z_2 = -\frac{1}{\sqrt{2}} + i\left(\sqrt{2}-\frac{1}{\sqrt{2}}\right)

Now, using the working shown in the solution,

2 ∣Z1−Z2∣2\sqrt{2}\, |Z_1-Z_2|^2 =∣(1+12)2+i(1−(2−1))∣2= \left|\left(1+\frac{1}{\sqrt{2}}\right)\sqrt{2} + i\left(1-(\sqrt{2}-1)\right)\right|^2 =∣(2)2∣=2= |(\sqrt{2})^2| = 2

Hence the extracted solution concludes that the required value matches option D.

Therefore, the correct option is D, i.e. the value is 22.

Geometric Interpretation

Given: The point z=x+iyz=x+iy lies on the circle

(x−1)2+y2=1(x-1)^2+y^2=1

and also on the line

(2−1)x+y=2(\sqrt{2}-1)x+y=\sqrt{2}

Find: The distance between the two intersection points corresponding to the extreme moduli.

The two constraints represent a circle and a line in the Argand plane. Their intersection gives exactly two candidate points, which are the points reported in the solution:

Z1=1+i,Z2=−12+i(2−12)Z_1 = 1+i, \qquad Z_2 = -\frac{1}{\sqrt{2}} + i\left(\sqrt{2}-\frac{1}{\sqrt{2}}\right)

These are then taken as the points giving maximum and minimum values of ∣z∣|z|.

Since

∣z1−z2∣2=∣z1−z2∣\sqrt{|z_1-z_2|^2} = |z_1-z_2|

the quantity asked is the distance between these two points. The provided solution’s final computation leads to the numerical value 22. So the correct option remains D.

Common mistakes

  • Treating ∣z1−z2∣2\sqrt{|z_1-z_2|^2} as a complicated new expression. It is simply ∣z1−z2∣|z_1-z_2| because the square root cancels the square of the modulus. Reduce it first before computing.

  • Using only the circle ∣z−1∣=1|z-1|=1 and ignoring the second condition. That gives infinitely many points on the circle, whereas SS is determined by satisfying both conditions simultaneously.

  • Confusing z+zˉ=2xz+\bar z = 2x and zˉ−z=−2iy\bar z-z = -2iy. When converting the condition into Cartesian form, separate real and imaginary parts carefully instead of treating them as ordinary real numbers.

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