Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 11 / c / Solution

For real $\lambda >0$, squaring the two nonnegative moduli gives $|z-i{|}^2={\lambda }^2|z+i{|}^2$. In Cartesian coordinates this isAssume $\lambda \ne 1$ and put $h=(1+{\lambda }^2)/(1-{\lambda }^2)$. Dividing by $1-{\lambda }^2$ yields $x^2+y^2-2hy+1=0$. Since $z{z}^{\ast }=x^2+y^2$ and $b{z}^{\ast }+b^{\ast }z=2\mathrm{Re}(b{z}^{\ast })$, the requested complex equation hasIndeed these choices make the mixed term $-2hy$. Completing the square givesThe Circles of Apollonius therefore haveThe 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.