Let $S=\{z \in C:|z-1|=1$ and $(\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2}\}$. Let $z_{1}, z_{2} \in S$ be such that $\left|z_{1}\right|=\max _{z \in S}|z|$ and $\left|z_{2}\right|=\min _{z \in S}|z|$. Then $\left|\sqrt{2} z_{1}-z_{2}\right|^{2}$ equals :

Complex Numbers — JEE Main · 2024 · 01 Feb 2024 · Shift 1 Question with Solution

Let

S = {z \in C : ∣ z - 1∣ = 1

and

(2 - 1) (z + \overset{z}{ˉ}) - i (z - \overset{z}{ˉ}) = 22}

. Let

z_{1}, z_{2} \in S

be such that

∣ z_{1} ∣ = max_{z \in S} ∣ z ∣

and

∣ z_{2} ∣ = min_{z \in S} ∣ z ∣

. Then

2 z_{1} - z_{2}^{2}

equals :

Answer: (D)

Write

z = x + i y

. From

∣ z - 1∣ = 1

(x - 1)^{2} + y^{2} = 1

…(1).
For the second condition, use

z + \overset{z}{ˉ} = 2 x

and

z - \overset{z}{ˉ} = 2 i y

(2 - 1) (2 x) - i (2 i y) = 22 ⟹ 2 (2 - 1) x + 2 y = 22 ⟹ (2 - 1) x + y = 2 \dots (2)

From (2):

y = 2 - (2 - 1) x

. Substituting into (1) and simplifying (with

k = 2 - 1

k^{2} = 3 - 22

x^{2} (4 - 22) - x (6 - 22) + 2 = 0 ⟹ x^{2} (2 - 2) - x (3 - 2) + 1 = 0

The discriminant is

(3 - 2)^{2} - 4 (2 - 2) = 11 - 62 - 8 + 42 = 3 - 22 = (2 - 1)^{2}

, so

x = \frac{( 3 - 2 ) \pm ( 2 - 1 )}{2 ( 2 - 2 )} ⟹ x = 1 or x = 1 + \frac{2}{2}

For

x = 1

y = 2 - (2 - 1) = 1

, giving

z_{a} = 1 + i

∣ z_{a} ∣ = 2

.
For

x = 1 + \frac{2}{2}

y = 2 /2

, giving

z_{b} = (1 + \frac{2}{2}) + i \frac{2}{2}

, and

∣ z_{b} ∣^{2} = (1 + \frac{2}{2})^{2} + \frac{1}{2} = 2 + 2

, so

∣ z_{b} ∣ = 2 + 2 > 2 = ∣ z_{a} ∣

.
So

z_{1} = z_{b}

(maximum modulus) and

z_{2} = z_{a} = 1 + i

(minimum modulus). Then

2 z_{1} - z_{2} = 2 [(1 + \frac{2}{2}) + i \frac{2}{2}] - (1 + i) = (2 + 1 + i) - (1 + i) = 2

∣ 2 z_{1} - z_{2} ∣^{2} = (2)^{2} = 2

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Single correct option