If $\frac{z-\alpha}{z+\alpha} \left(\alpha\in R\right) $ is a purely imaginary number and $|z| = 2$, then a value of $\alpha$ is :
- $1$
- $2$
- $\sqrt{2}$
- $\frac{1}{2}$
The Correct Option is B
Solution and Explanation
$ z\bar{z}+z\alpha -\alpha\bar{z} -\alpha^{2} +z\bar{ z } - z\alpha + \bar{z} \alpha-\alpha^{2} = 0$
$ \left|z\right|^{2} = \alpha^{2} , a = \pm2 $
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
