Consider the set of 4-tuples of complex numbers $(x_1,x_2,y_1,y_2)$ for which $$\Im\left( -\frac{-y_1 x_1-y_2 x_2}{x_1^2+x_2^2} \right)=0 \tag{1}$$ and $$\Im\left(\frac{i (y_1 x_2-y_2 x_1)}{x_1^2+x_2^2}\right)=0 \tag{2}$$ (1) and (2) are equivalent to the condition that there exist some real $a,b$ such that $$ \left[\begin{matrix} a & i b \\ - i b & a \end{matrix}\right] \left[\begin{matrix} x_1\\ x_2 \end{matrix}\right]=\left[\begin{matrix} y_1\\ y_2 \end{matrix}\right] \tag{3} $$ (1) and (2) are what you get by solving (3) for $a,b$ and demanding they are real.
This is a somewhat open-ended question: I hope to find some sort of vector space or group structure on this set, (or even just a binary operation), something that would allow me to find a third 4-tuple that obeys (1) and (2), given two 4-tuples that obey (1) and (2).
Some thoughts: Clearly, if $(x_1,x_2,y_1,y_2)$ obeys (1) and (2), then so does any multiple $(r x_1,r x_2,r y_1,r y_2)$ for any complex $r$. So, a useful reformulation might be $$ (r \cos(\theta),r \sin(\theta),r y_1,r y_2)$$ with both $r$ and $\theta$ complex. Then (1) and (2) reduce to $$\Im\left( y_1 \cos(\theta)+y_2 \sin(\theta) \right)=0 $$ $$\Im\left( [y_1,y_2]\cdot [\cos(\theta),\sin(\theta)] \right)=0 $$ and $$\Re\left(-y_1 \sin(\theta)+y_2 \cos(\theta)\right)=0 $$ $$\Re\left( [y_1,y_2]\cdot [-\sin(\theta),\cos(\theta)] \right)=0 $$