Some equation with complex numbers Given $a,b \in \mathbb{C}$ such that $a^2+b^2=1$, it is clear that $x:=a\bar{a}+b\bar{b}$ is a real number and that $yi:=a\bar{b}-\bar{a}b$ is imaginary (i.e $y$ is real). Moreover, a direct computation shows that $x,y$ satisfy $x^2-y^2=1$. 
Now, the question is whether the converse holds as well. Namely, given $x,y\in \mathbb {R}$ such that $x^2-y^2=1$, are there $a,b\in \mathbb{C}$ with $a^2+b^2=1$ and such that $x=a\bar{a}+b\bar{b}$ and $yi=a\bar{b}-\bar{a}b$?
Unfortunatly, the motivation for this is a bit difficult to explain, so I will not try to.

The numbers $x,y$ need to satisfy $x\geq1$, $y^2=x^2-1$. 
Note that 
$$
x-y=|a|^2+|b|^2-\frac{a\bar{b}-\bar{a}b}i=|a|^2+|b|^2+i(a\bar{b}-\bar{a}b)=|a-ib|^2.
$$
Similarly, 
$$
x+y=|a+ib|^2.
$$
If $a+ib$ and $a-ib$ are real (they don't have to in principle, but let's assume; we are choosing to have $a$ real and $b$ imaginary), then we would have
$$
a=\frac{\sqrt{x-y}+\sqrt{x+y}}2,\ \ b=\frac{\sqrt{x+y}-\sqrt{x-y}}{2i}.
$$
It is easy to check that $a^2+b^2=1$, $|a|^2+|b|^2=x$, $-i(a\bar{b}-\bar{a}b)=y$.
