Let C be the set of all complex numbers. Let $S_{1}=\left\{z \in C\||z-3-2 i|^{2}=8\right\}$, $S_{2}=\{z \in C \mid \operatorname{Re}(z) \geq 5\}$ and $S_{3}=\{z \in C \mid |z-\bar{z}| \geq 8\}$. Then the number of element in $\mathrm{S}_{1} \cap \mathrm{S}_{2} \cap \mathrm{S}_{3}$ is equal to:

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Complex Numbers

Let C be the set of all complex numbers. Let

S_{1} = {z \in C ∥∣ z - 3 - 2 i ∣^{2} = 8}

S_{2} = {z \in C ∣ Re (z) \geq 5}

and

S_{3} = {z \in C ∣ ∣ z - \overset{z}{ˉ} ∣ \geq 8}

.
Then the number of element in

S_{1} \cap S_{2} \cap S_{3}

is equal to:

Answer: (C)

S_{1}

is the circle

(x - 3)^{2} + (y - 2)^{2} = 8

. With

z = x + i y

∣ z - \overset{z}{ˉ} ∣ = ∣2 i y ∣ = 2∣ y ∣

, so

S_{3}

∣ y ∣ \geq 4

, while

S_{2}

x \geq 5

.
On the circle,

x \geq 5

forces

(x - 3)^{2} \geq 4

, hence

(y - 2)^{2} \leq 4

, that is

0 \leq y \leq 4

. Together with

∣ y ∣ \geq 4

this leaves

y = 4

, and then

(x - 3)^{2} = 8 - 4 = 4

gives

x = 5

x = 1

, of which only

x = 5

satisfies

x \geq 5

.
The intersection is the single point

z = 5 + 4 i

, so it has

1

element.

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option