Question

The region represented by z such that $\left|\frac{z-a}{z+a}\right|=1\left(Im\right(a)=0)$ is

• A. $y=0$
• B. $x=0$
• C. $x+y=0$
• D. $x-y=0$

Answer 1

The correct option is B $x=0$

Let $z=x+iy$

$\Rightarrow \left|\right(x-a)+iy|=\left|\right(x+a)+iy|$

$\Rightarrow (x-a{)}^2+\text{/}{y}^2=(x+a{)}^2+\text{/}{y}^2$

$\Rightarrow \text{/}{x}^2+\text{/}{a}^2-2ax=\text{/}{x}^2+\text{/}{a}^2+2ax$

$\Rightarrow 0$