S1:∣z−3−2i∣2=8
⇒∣z−3−2i∣=22
⇒(x−3)2+(y−2)2=(22)2 (Put z=x+iy)
S2:x≥5
S3:∣z−zˉ∣≥8 (put z=x+iy)
⇒∣2iy∣≥8
⇒2∣y∣≥8∴y≥4,y≤−4
So, the required diagram is

Hence, n(S1∩S2∩S3)=1
JEE Main 2021 — Mathematics Algebra
Let C be the set of all complex numbers. Let
S1=z∈C∣∣z–3–2i∣2=8,
S2=z∈C∣Re(z)≥5 and
S3=z∈C∣∣z–zˉ∣≥8.
Then the number of elements in S1∩S2∩S3 is equal to
Held on 27 Jul 2021 · Verified 6 Jul 2026.
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