Let S _ 1 , S _ 2 and S _ 3 be three sets defined as S _ 1 = z ∈ ℂ: | z - 1 | ≤ 2 , S _ 2 = z ∈ ℂ: Re ( ( 1 - i ) z ) ≥ 1 and S _ 3 = z ∈ ℂ: Im ( z ) ≤ 1 . Then, the set S _ 1 S _ 2 S _ 3

JEE Main 202117 MarchEvening Shift

Exam: JEE Main
Year: 2021
Date: 17 March
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2021 · 17 March · Evening Shift · Mathematics · Complex Number

Let S _ 1 , S _ 2 and S _ 3 be three sets defined as S _ 1 = z ∈ ℂ:...

Let ${S}_{1}, {S}_{2}$ and ${S}_{3}$ be three sets defined as

${S}_{1} = { z \in \mathbb{C}: | z - 1 | \le \sqrt{2} } ,$

${S}_{2} = { z \in \mathbb{C}: Re ( ( 1 - i ) z ) \ge 1 }$ and

${S}_{3} = { z \in \mathbb{C}: Im ( z ) \le 1 } .$

Then, the set ${S}_{1} \cap {S}_{2} \cap {S}_{3}$

Let S _ 1 , S _ 2 and S _ 3 be three sets defined as S _ 1 = z ∈ ℂ: | z - 1 | ≤ 2 , S _ 2 = z ∈ ℂ: Re ( ( 1 - i ) z ) ≥

Options

Ais a singleton
Bhas exactly two elements
Chas infinitely many elements
Dhas exactly three elements

Correct answer

C. has infinitely many elements

Step-by-step solution

We know that the complex number $z$ satisfying \left|z - {z}_{0}\right| = r $represents a circle with centre$ {z}_{0} $and radius$ r $units.Hence, for$ {S}_{1} = | z - 1 | \le \sqrt{2}, z $lies on and inside the circle of radius$ \sqrt{2} $units and centre$ ( 1, 0 ) .

For ${S}_{2},$ let $z = x + iy$

Now, $( 1 - i ) ( z ) = ( 1 - i ) ( x + iy )$

$⇒ ( 1 - i ) ( z ) = x + iy - ix - {i}^{2}y$

$⇒ ( 1 - i ) ( z ) = x + iy - ix + y$

$⇒ Re ( ( 1 - i ) z ) = x + y$

$⇒ x + y \ge 1 .And,$ for {S}_{3}, $again let$ z = x + iy, $$ ⇒ y \le 1 .

Plotting all $the inequalities on the graph, we get<img alt="" src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/c3baaf5f-a18c-4c55-865f-843c5061760e-a60846ca-993c-418c-bfdb-b8506fdba3e3-image.png" style="width: 300px; height: 195px;" />Now, the common part is shown in the shaded part, hence, we get infinite points in the shaded region.⇒ {S}_{1} \cap {S}_{2} \cap {S}_{3}$ has infinitely many elements.

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