Let C be the set of all complex numbers. Let S 1 = z ∈ C |   z – 3 – 2 i 2 = 8 , S 2…

Let C be the set of all complex numbers. Let

S1=z∈C| z–3–2i2=8,

S2=z∈C| Rez≥5  and

S3=z∈C| z–z¯≥8.

Then the number of elements in S1∩S2∩S3 is equal to

  1. 1
  2. 0
  3. 2
  4. Infinite

Solution

S1:z-3-2i2=8

⇒z-3-2i=22

⇒x-32+y-22=222 (Put z=x+iy)

S2:x≥5

S3:z-z¯≥8 (put z=x+iy)

⇒2iy≥8

⇒2y≥8  ∴ y≥4,y≤-4

So, the required diagram is 

Hence, nS1∩S2∩S3=1

Asked in: JEE Main 2021 (27 Jul Shift 1)

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