Complex Numbers And Quadratic Equations question 58
Question: If z is a complex number such that $ \frac{z-1}{z+1} $ is purely imaginary, then [MP PET 1998, 2002]
Options:
A) $ |z|=0 $
B) $ |z|=1 $
C) $ |z|>1 $
D) $ |z|<1 $
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ \frac{z-1}{z+1}=iy $ where $ y\in R $ This gives $ z=\frac{1+iy}{1-iy}=\frac{1+iy}{1-iy}\times \frac{1+iy}{1+iy}=\frac{(1-y^{2})+2iy}{1+y^{2}} $
$ \therefore $ $ |z|=\frac{1}{1+y^{2}}\sqrt{{{(1-y^{2})}^{2}}+4y^{2}}=\frac{1+y^{2}}{1+y^{2}}=1 $ .
BETA
