Find non-zero real numbers $a,b,c,d$ such that $a^2+c^2=b^2+d^2$ and $ab+cd=0$.

Question

There are the obvious cases where $a=c=0$ and $b=d$. Are there any cases for this where all four numbers are non-zero?

Note, this problem is equivalent to saying the matrix $A^{-1}JA$ is a rotation invariant linear complex structure where $A=\begin{pmatrix}a & b\\c & d\end{pmatrix}\in GL(2,\mathbb{R})$ and $J=\begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}$ is the usual linear complex structure.

Answer 1

$$d=-\frac{ab}{c}$$ and we obtain: $$a^2+c^2=b^2+\frac{a^2b^2}{c^2}$$ or $$(a^2+c^2)(b^2-c^2)=0.$$

If $b=c$ then we get $d=-a$.

if $b=-c$ then we get $d=a$.

Id est, we got the answer: $$\{(a,b,b,-a),(a,b,-b,a)|a\neq0,b\neq0\}$$

Answer 2

Consider $v=(a,c)$ and $w=(b,d)$. Then you want that $v$ and $w$ be of the same length, and that they be perpendicular.

Once you fix $v$, which you can think of as having a parameter of length $r$ and angle $\theta$ with the $x$-axis, $w$ is determined as any of the two points in the line perpendicular to $v$ that cross the circle of radius $r$ at the origin.

You can think of the space parametrizing your solution set as a union of $\mathbb R_+\times S^1\times S^0$. The first factor gives you the length, the second the angle, and the third the choice of direction.