Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

2 Answers

up vote 2 down vote accepted

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.)

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.