= Solution
Since $z^2=(x^2-y^2)+2ixy$, equating its <real part> and <imaginary part> gives $x^2-y^2=2xy$, or
$$
y^2+2xy-x^2=0.
$$
The <quadratic equation> in $y$ factors as $[y-(-1+\sqrt2)x][y-(-1-\sqrt2)x]=0$. Thus the set is the union of two <straight lines> through the origin:
$$
\boxed{y=(-1+\sqrt2)x\quad\text{or}\quad y=(-1-\sqrt2)x.}
$$
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.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1-q11-loci.png]
{title=The two pairs of lines in the Argand diagrams of Question 11}
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