JEE MAIN - Mathematics (2019 - 12th January Morning Slot - No. 4)
If $${{z - \alpha } \over {z + \alpha }}\left( {\alpha \in R} \right)$$ is a purely imaginary number and | z | = 2, then a value of $$\alpha $$ is :
$${1 \over 2}$$
$$\sqrt 2 $$
2
1
Explanation
$${{z - \alpha } \over {z + \alpha }} + {{\overline z - \alpha } \over {\overline z + \alpha }} = 0$$
$$z\overline z + z\alpha - \alpha \overline z - {\alpha ^2} + z\overline z - z\alpha + \overline z \alpha - {\alpha ^2} = 0$$
$${\left| z \right|^2} = {\alpha ^2},$$ $$a = \pm 2$$
$$z\overline z + z\alpha - \alpha \overline z - {\alpha ^2} + z\overline z - z\alpha + \overline z \alpha - {\alpha ^2} = 0$$
$${\left| z \right|^2} = {\alpha ^2},$$ $$a = \pm 2$$
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