Solution
= Solution
The original PDF places a <complex modulus> around the quotient; those bars are essential. Its denominator excludes $z=-i$. Using $|z^*-i|=|z+i|$, the case $\lambda=1$ becomes $|z-i|=|z+i|$. In Cartesian coordinates,
$$
x^2+(y-1)^2=x^2+(y+1)^2\quad\Longrightarrow\quad\boxed{y=0.}
$$
This <locus> is the real axis, the perpendicular bisector of the points $i$ and $-i$ in the <complex plane>. The excluded denominator point does not lie on it.