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$\begin{align*}\frac{|z-a|}{|z+a|} &= 1\\|z-a|&= |z+a|\\|z-a|^2&= |z+a|^2\\|x + iy - a|^2&= |x + iy + a|^2\\( x-a)^2 + y^2&= ( x+a)^2 + ^2\\  x^2 - 2ax + a^2&=  x^2 + 2ax + a^2\\ x &= 0 \end{align*}$

So, it represents $y$ axis and the answer is B.

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