Let $S_{1}, S_{2}$ and $S_{3}$ be three sets defined as $S_{1}=\{z \in C:|z-1| \leq \sqrt{2}\}$ $S_{2}=\{z \in C: \operatorname{Re}((1-i) z) \geq 1\}$ $S_{3}=\{z \in C: \operatorname{Im}(z) \leq 1\}$ Then the set $S_{1} \cap S_{2} \cap S_{3}$

(C) Has infinitely many elements

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Complex Numbers

Let

S_{1}, S_{2}

and

S_{3}

be three sets defined as

S_{1} = {z \in C : ∣ z - 1∣ \leq 2}

S_{2} = {z \in C : Re ((1 - i) z) \geq 1}

S_{3} = {z \in C : Im (z) \leq 1}

Then the set

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

Answer: (C)

Put

z = x + i y

. Then

S_{1}

is the closed disc

(x - 1)^{2} + y^{2} \leq 2

; since

Re ((1 - i) z) = x + y

, the set

S_{2}

is the half-plane

x + y \geq 1

; and

S_{3}

is the half-plane

y \leq 1

.
The centre

(1, 0)

of the disc satisfies

x + y = 1

and

y = 0

, so it lies in all three sets. Take any point

(1 + h, 0)

with

0 < h < 2

: it satisfies

x + y = 1 + h > 1

y = 0 < 1

and

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

, so it too lies in all three sets.
The intersection contains a whole segment of points, so it has infinitely many elements.

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option