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JEE Main 202228 JulyMorning Shift
Exam
JEE Main
Year
2022
Date
28 July
Shift
Morning Shift
Subject
Mathematics
Chapter
Complex Number

JEE Main 2022 · 28 July · Morning Shift · Mathematics · Complex Number

Let S _ 1 = \ z _ 1 ∈ C: | z _ 1 - 3| = (1)/(2)\ and S _ 2 = \ z _...

Let S1={z1∈C:∣z1−3∣=12}and{S}_{1} = \left\{{z}_{1} \in C: \left|{z}_{1} - 3\right| = \dfrac{1}{2}\right\} and{S}_{2} = \left\{{z}_{2} \in C: \left|{z}_{2} - \left|{z}_{2} + 1\right|\right| = \left|{z}_{2} + \left|{z}_{2} - 1\right|\right|\right\} .Then,forThen, for{z}_{1} \in {S}_{1}andand{z}_{2} \in {S}_{2},theleastvalueof∣z2−z1∣, the least value of \left|{z}_{2} - {z}_{1}\right| isLet S _ 1 = \ z _ 1 ∈ C: | z _ 1 - 3| = (1)/(2)\ and S _ 2 = \ z _ 2 ∈ C: | z _ 2 - | z _ 2 + 1|| = | z _ 2 + | z _ 2 -

Options

  1. A00
  2. B12\dfrac{1}{2}
  3. C32\dfrac{3}{2}
  4. D52\dfrac{5}{2}

Correct answer

C. 32\dfrac{3}{2}

Step-by-step solution

Here ∣z1−3∣=12\left|{z}_{1} - 3\right| = \dfrac{1}{2} represents a circle on argand plane with centre (3,0)\left(3, 0\right) and radius 12\dfrac{1}{2}

Given ∣z2+∣z2−1∣∣2=∣z2−∣z2+1∣∣2{\left|{z}_{2} + \left|{z}_{2} - 1\right|\right|}^{2} = {\left|{z}_{2} - \left|{z}_{2} + 1\right|\right|}^{2}

⇒(z2+∣z2−1∣)(z¯2+∣z2−1∣)=(z2−∣z2+1∣)(z¯2−∣z2+1∣)⇒ \left({z}_{2} + \left|{z}_{2} - 1\right|\right)\left(\overset{¯}{z}2 + \left|{z}_{2} - 1\right|\right) = \left({z}_{2} - \left|{z}_{2} + 1\right|\right)\left(\overset{¯}{z}2 - \left|{z}_{2} + 1\right|\right)

⇒z2(∣z2−1∣+∣z2+1∣)+z¯2(∣z2−1∣+∣z2+1∣)=∣z2+1∣2−∣z2−1∣2⇒ {z}_{2}\left(\left|{z}_{2} - 1\right| + \left|{z}_{2} + 1\right|\right) + \overset{¯}{z}2\left(\left|{z}_{2} - 1\right| + \left|{z}_{2} + 1\right|\right) = {\left|{z}_{2} + 1\right|}^{2} - {\left|{z}_{2} - 1\right|}^{2}

⇒(z2+z¯2)(∣z2+1∣+∣z2−1∣)=2(z2+z¯2)⇒ \left({z}_{2} + \overset{¯}{z}2\right)\left(\left|{z}_{2} + 1\right| + \left|{z}_{2} - 1\right|\right) = 2\left({z}_{2} + \overset{¯}{z}2\right)

⇒⇒ Either z2+z¯2=0{z}_{2} + \overset{¯}{z}2 = 0 or \left|{z}_{2} + 1\right| + \left|{z}_{2} - 1\right| = 2</p><p>i.e.</p><p>i.e.{z}_{2}liesonimaginaryaxisoritliesonthelinesegmentjoining(−1,0)lies on imaginary axis or it lies on the line segment joining \left(- 1, 0\right) and \left(1, 0\right)</p><p><imgsrc="https://cdn−question−pool.getmarks.app/pyq/jeemain/cfa52ac5−9902−4fd6−99c7−2e86e16274ed−image.png"style="width:315px;height:200px;"/></p><p>So,theminimumdistancebetween</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/cfa52ac5-9902-4fd6-99c7-2e86e16274ed-image.png" style="width: 315px; height: 200px;" /></p><p>So, the minimum distance between{z}_{1} \text{ and } {z}_{2}willbethedistancebetweenthepoints(1,0) and (52,0)will be the distance between the points \left(1, 0\right) \text{ and } \left(\dfrac{5}{2}, 0\right)

Hence, {\left|{z}_{1} - {z}_{2}\right|}_\min = \dfrac{3}{2}

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