This is an old qual problem at my school. I've never dealt with quadratic forms, so I'm not sure if I've done this right.
Two quadratic forms $Q_1(x,y)$ and $Q_2(x',y')$ are said to be equivalent if they are related by a non-singular change of coordinates $(x,y)\mapsto (x',y')$. Decide whether $Q_1=xy$ and $Q_2=x^2+y^2$ are equivalent over $\mathbb{C}$ and whether they are equivalent over $\mathbb{R}$.
I'm not sure what "related by" actually means. I take it to mean there is a transformation $x=ax'+by'$ and $y=cx'+dy'$ such that $ad-bc\neq 0$. My guess is $Q_1$ and $Q_2$ are equivalent over $\mathbb{C}$, since you can take the change of coordinates $x=x'+iy'$ and $y=x'-iy'$. So $xy=(x'+iy')(x'-iy')=x'^2+y'^2$, and the change of coordinate matrix would be $$ \begin{bmatrix} 1 & 1 \\ i & -i \end{bmatrix} $$ which is nonsingular with determinant $-2i$.
However, they are not equivalent over $\mathbb{R}$. If there existed a transformation $x=ax'+by'$ and $y=cx'+dy'$ such that $xy=x'^2+y'^2$, we would have to have $$ xy=acx'^2+(ad+bc)x'y'+bdy'^2 $$ with $ac=1$, $bd=1$ and $ad+bc=0$. Writing $c=1/a$ and $d=1/b$, the last equation simplifies to $$ \frac{a}{b}+\frac{b}{a}=0\implies a^2+b^2=0 $$ for nonzero $a,b\in\mathbb{R}$, which is impossible.
Did I interpret this correctly?
Yes, this is correct. Another way to see the inequivalence over $\mathbb R$ is to denote that (as $x$ and $y$ range over various elements of $\mathbb R$), the quadratic form $xy$ can take both negative as well as positive values, while $x^2 + y^2$ takes only non-negative values.