If \(\frac{z-\alpha }{z+\alpha }\) (α ∈ R) is a purely imaginary number and |z| = 2, then a value of α is number and |z| = 2, then a value of α is

If \(\frac{z-\alpha }{z+\alpha }\) (α ∈ R) is a purely imaginary

1.	If \(\frac{z-\alpha }{z+\alpha }\) (α ∈ R) is a purely imaginary number and \|z\| = 2, then a value of α is number and \|z\| = 2, then a value of α is
A.	√2
B.	1/2
C.	1
D.	2
Answer» E.