Question:

 If z=(11+2i−5i1−2i−35+3i5i5−3i7) then (i=−1)

Updated On: Apr 27, 2024
  • (A) z is purely real 
  • (B) z is purely imaginary
  • (C) z+z¯=0
  • (D) (z−z¯) i is purely imaginary
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The Correct Option is A

Solution and Explanation

Explanation:
 If z=(11+2i−5i1−2i−35+3i5i5−3i7) then (i=−1)
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