JEE Main 12 Jan 2019 Morning Question 7
Question: If $ \frac{z-\alpha }{z+\alpha }(\alpha \in R) $ is a purely imaginary number and $ |z|=2, $ then a value of $ \alpha $ is [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) 1
B) $ \sqrt{2} $
C) $ \frac{1}{2} $
D) 2
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ z=x+iy $
$ \therefore $ $ |z|=\sqrt{x^{2}+y^{2}}=2\Rightarrow x^{2}+y^{2}=4 $
Now, $ \frac{z-\alpha }{z+\alpha }=\frac{x+iy-\alpha }{x+iy+\alpha }=\frac{(x-\alpha )+iy}{(x+\alpha )+iy}\times \frac{(x+\alpha )-iy}{(x+\alpha )-iy} $ $ =\frac{(x^{2}+y^{2}-{{\alpha }^{2}})}{{{(x+\alpha )}^{2}}+y^{2}}\times \frac{i2\alpha y}{{{(x+\alpha )}^{2}}+y^{2}} $
$ \Rightarrow $ $ \frac{x^{2}+y^{2}-{{\alpha }^{2}}}{{{(x+\alpha )}^{2}}+y^{2}}=0 $ $ [ \because \frac{z-\alpha }{z+\alpha } $ is purely imaginary]
$ \Rightarrow $ $ x^{2}+y^{2}-{{\alpha }^{2}}=0 $
$ \Rightarrow $ $ {{\alpha }^{2}}=4 $
$ \Rightarrow $ $ \alpha =\pm 2 $
