Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 11 / b / i / Solution

Since $z^2=(x^2-y^2)+2ixy$, equating its real part and imaginary part gives $x^2-y^2=2xy$, orThe quadratic equation in $y$ factors as $[y-(-1+\sqrt{2})x][y-(-1-\sqrt{2})x]=0$. Thus the set is the union of two straight lines through the origin:Their slopes multiply to $-1$, so they are perpendicular. Their angles to the real axis are $\pi /8$ and $-3\pi /8$, modulo $\pi$. The origin is included.