JEE Main 2022MathematicsComplex NumberMediumMCQ

JEE Main 2022 — Complex Number Question with Solution

JEE Main 2022 (28 Jul Shift 1)

Question

Let S1=z1∈C:z1-3=12 and S2=z2∈C:z2-z2+1=z2+z2-1. Then, for z1∈S1 and z2∈S2, the least value of z2-z1 is

Choose an option

Show full solutionCorrect option: C
Correct answer
C32

Step-by-step explanation

Here z1-3=12 represents a circle on argand plane with centre 3,0 and radius 12

Given z2+z2−12=z2−z2+12

⇒z2+z2-1z¯2+z2-1=z2-z2+1z¯2-z2+1

⇒z2z2-1+z2+1+z¯2z2-1+z2+1=z2+12-z2-12

⇒z2+z¯2z2+1+z2-1=2z2+z¯2

⇒ Either z2+z¯2=0 or z2+1+z2-1=2

i.e. z2 lies on imaginary axis or it lies on the line segment joining -1,0 and 1,0

So, the minimum distance between z1 & z2 will be the distance between the points 1,0 & 52,0

Hence, z1-z2min=32

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About this question

This is a previous-year question from JEE Main 2022, covering the Complex Number chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.