Topic: Engineering Mathematics
The region of the z plane for which $|\frac{z-a}{z+a}|=1(Re,a\ne0)is-$
- A. x-axis
- B. y-axis
- C. The straight line $z=|a|$
- D. None of the above
Answer: B. y-axis
Explanation: Given $|\frac{z-a}{z+a}|=1 \Rightarrow |z-a| = |z+a| \Rightarrow |z-a|^{2} = |z+a|^{2}$. Let $z = x+iy$. Then $|x+iy-a|^{2} = |x+iy+a|^{2} \Rightarrow (x-a)^{2}+y^{2} = (x+a)^{2}+y^{2} \Rightarrow x^{2}+a^{2}-2ax = x^{2}+a^{2}+2ax \Rightarrow 4ax=0$. Since $a \neq 0$, it must be that $x=0$, which represents the y-axis.