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if z is a complex number the number of solutions of z²+conjugate(z)=0


  1. if z is a complex number the number of solutions of

  2. z²+conjugate(z)=0

Grade:12th pass

3 Answers

shubham sharma
27 Points
8 years ago
there will be three solutions
Amit Jain
askIITians Faculty 32 Points
8 years ago
use z= x+iy and solve the equation. You will get three roots: (0+0i), (0.5+iroot3/2), (0.5-iroot3/2). Hope that helps. Amit Jain AskIITians Faculty
vamsi
43 Points
7 years ago
3 SOLUTIONS

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