InterviewSolution
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InterviewSolution
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The number of solutions of the equation `z^2+z=0` where z is a a complex number, isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - D Let `z = x + iy`, so that `barz = x -iy`. `therefore z ^(2) + barz = 0` `rArr (x^(2) -y^(2) + x) + i(2xy - y) = 0` Equating real and imaginary parts, we get `x ^(2)- y^(2) + x = 0` and `2xy - y = 0 rArr y = 0 or x = (1)/(2)` If `y 0`, then (1) given `x^(2) + x= 0` `rArr x = 0 or x = -1` If `x = 1//2`, then from Eq.(1) `y^(2) = (1)/(4) + (1)/(2) = (3)/(4) or y = pm (sqrt(3))/(2)` Hence, there are four solutions in all. |
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