JEE Mains · Maths · STD 11 - 4.1 complex nubers
If \(\frac{{z - \alpha }}{{z + \alpha }}\left( {\alpha \in R} \right)\) is a purely imaginary number and \(\left| z \right| = 2\), then a value of \(\alpha \) is
- A \(2\)
- B \(1\)
- C \(\frac{1}{2}\)
- D \(\sqrt 2\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
\(\frac{{z\, - \,\alpha }}{{z\, + \,\alpha }}\, = \, - \,\left( {\frac{{\bar z\, - \,\alpha }}{{\bar z\, + \,\alpha }}} \right)\) \(z\bar z\, + \,z\alpha \, - \,\alpha \bar z\, - \,{\alpha ^2}\) \( = \, - \,z\bar z\, + \,\,\alpha \bar z\, - \,\alpha \bar z\, + {\alpha ^2}\,\)…
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