Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2020 / Paper 1 / 11 / b

Write $z=x+iy$. Then $|z^{\ast }+2i|=\sqrt{x^2+(y-2{)}^2}$ and $|z|=\sqrt{x^2+y^2}$, so the condition fixes the sum of distances from $(0,2)$ and $(0,0)$. It is an ellipse with centre $(0,1)$, vertical semimajor axis $A=2$, and focal distance $c=1$. Its horizontal semiminor axis obeys $B^2=A^2-c^2=3$. Thus the locus isThe vertices are $(0,-1)$ and $(0,3)$; the horizontal extremities are $(\pm \sqrt{3},1)$. The complex conjugation places the second focus above the origin.

The triangle inequality bounds the distance sum below by the separation of the two foci, which is two. Equality holds exactly for points on the closed segment joining them. Hence the locus isBoth endpoints are included. It is the segment on the positive imaginary axis shown in the right-hand panel above.