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

The original PDF places a complex modulus around the quotient; those bars are essential. Its denominator excludes $z=-i$. Using $|z^{\ast }-i|=|z+i|$, the case $\lambda =1$ becomes $|z-i|=|z+i|$. In Cartesian coordinates,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.

A quotient with nonzero denominator has zero complex modulus exactly when its numerator is zero. ThusAt that point $z^{\ast }-i=-2i\ne 0$, so the solution is admissible.

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.

For arbitrary complex $b$ and real $c$, complete the complex modulus square:The locus is a single point exactly when

The same completed square gives $|z+b{|}^2=|b{|}^2-c$. A nondegenerate circle therefore occurs exactly whenFor completeness, $c>|b{|}^2$ gives an empty locus because a squared complex modulus cannot be negative.

No finite choice of $b$ and $c$ gives a line. The coefficient of $z{z}^{\ast }=x^2+y^2$ is fixed at one, so completing the square always gives a circle, a point, or the empty set. A general equation $Az{z}^{\ast }+b{z}^{\ast }+b^{\ast }z+c=0$ could describe a line when $A=0$ and $b\ne 0$, but changing the fixed quadratic coefficient is not permitted here. A limiting sequence of increasingly large circles does not make any one of these finite equations a line.