Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 11 / b

Completing the square directly with the complex modulus givesFor $|b{|}^2>c$, this is the circle with centre $-b$ and radius $\sqrt{|b{|}^2-c}$. Equality gives just the point $-b$, and $|b{|}^2<c$ gives the empty locus. In particular, the nondegenerate answer is

Since $z\bar{z}=x^2+y^2$, the real quadratic equation isCompleting the square makes its geometry explicit:

A squared complex modulus is nonnegative. Conversely, every nonnegative right-hand side of $|z+b{|}^2=|b{|}^2-c$ can be attained. The necessary and sufficient condition is therefore

There is a single point exactly when the radius is zero:Then $|z+b{|}^2=0$ forces this unique value of the complex number $z$.