= Solution
Write $z=x+iy$. Then $|z^*+2i|=\sqrt{x^2+(y-2)^2}$ 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> is
$$
\boxed{\frac{x^2}{3}+\frac{(y-1)^2}{4}=1.}
$$
The vertices are $(0,-1)$ and $(0,3)$; the horizontal extremities are $(\pm\sqrt3,1)$. The <complex conjugation> places the second focus above the origin.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1-2020-q11-complex-loci.png]
{title=Distance-sum loci in the complex plane: an ellipse and its limiting segment}
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