= Solution
For real $\lambda>0$, squaring the two nonnegative moduli gives $|z-i|^2=\lambda^2|z+i|^2$. In Cartesian coordinates this is
$$
(1-\lambda^2)(x^2+y^2+1)-2(1+\lambda^2)y=0.
$$
Assume $\lambda\ne1$ and put $h=(1+\lambda^2)/(1-\lambda^2)$. Dividing by $1-\lambda^2$ yields $x^2+y^2-2hy+1=0$. Since $zz^*=x^2+y^2$ and $bz^*+b^*z=2\operatorname{Re}(bz^*)$, the requested complex equation has
$$
\boxed{b=-i\frac{1+\lambda^2}{1-\lambda^2},\qquad c=1.}
$$
Indeed these choices make the mixed term $-2hy$. Completing the square gives
$$
x^2+(y-h)^2=h^2-1=\frac{4\lambda^2}{(1-\lambda^2)^2}.
$$
The <Circles of Apollonius> therefore have
$$
\boxed{\text{centre }z_c=i\frac{1+\lambda^2}{1-\lambda^2},\qquad
\text{radius }\frac{2\lambda}{|1-\lambda^2|}.}
$$
The <radius> is positive for the stated range. No spurious point at $z=-i$ has been introduced: at that point the undivided equation would require $4=0$. Positive $\lambda<1$ gives a <circle> above the real axis, while $\lambda>1$ gives one below it.
Back to article page
