InterviewSolution
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InterviewSolution
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Let `(z-alpha)/(z+alpha)` is purely imaginary and `|z|=2, alphaepsilonR` then `alpha` is equal to (A) `2` (B) `1` (C) `sqrt2` (D) `sqrt3`A. `sqrt(2)`B. `1/2`C. 1D. 2 |
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Answer» Correct Answer - D Since the complex number `(z-a)/(z+a)(alpha ne R)`j is purely imaginary number therefore `(z-alpha)/(z+alpha)+(barz-alpha)/(bar z + alpha)=0 " "[ therefore alpha in R]` `rArr zbarz-abarz+az-alpha^2+zbarz-alphabarz-alpha^2=0` `rArr 2|z|^2-2alpha^2=0" "[because zbarz=|z|^2]` `rArr alpha^2=|z|^2=4` `rArr alpha=pm2` |
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