# 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.

-
Both answers are good. I will accept the first one published. – KotelKanim Oct 23 '12 at 22:52
I would upvote both of the answers, but I am having problems with logging in. Anyone who sees this, please upvote the answers :) – KotelKanim Oct 23 '12 at 22:53
Problems with logging in are resolved. +1 to both answers! – KotelKanim Mar 30 '13 at 12:00

## 2 Answers

You certainly need to assume $x\ge0$ (which then implies $x\ge1$), since you want $x=|a|^2+|b|^2$, which is always $\ge 0$. Once you have that, you can get your $a$ and $b$ with the idea that $a=\cos w$ and $b=\sin w$ with some $w=u+iv\in\mathbb{C}$. A bit of calculation then shows $x=\cosh 2v$ and $y=-\sinh 2v$. (I think I got the sign right, but you better check...) So you have to choose $v=-\frac12\sinh^{-1} y$, choose any $u\in \mathbb{R}$, and then $a=\cos w$ and $b=\sin w$ with $w=u+iv$ will do. The particular choice $u=0$ gives you $a=\cos( \frac{i}2\sinh^{-1} y) = \cosh(\frac12\sinh^{-1}y)$ and $b=-\sin(\frac{i}2\sinh^{-1} y) = -i\sinh(\frac12 \sinh^{-1} y)$. (These can undoubtedly be simplified.)

-

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 $a^2+b^2=1$, $|a|^2+|b|^2=x$, $-i(a\bar{b}-\bar{a}b)=y$.

-