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.

I want to ask for verification about whether this equation can be proven. If so, what is the best way to approach it? I tried this way... but I don't know how to continue on.

$$|B-Ae^{-j\omega\delta}|=|C-Ae^{+j\omega\delta}|$$ $$(B-Ae^{-j\omega\delta})(B^*-Ae^{+j\omega\delta})=(C-Ae^{+j\omega\delta})(C^*-Ae^{-j\omega\delta})$$ $$|B|^2-AB^*e^{-j\omega\delta}-ABe^{+j\omega\delta}+A^2 = |C|^2-AC^*e^{+j\omega\delta}-ACe^{-j\omega\delta}+A^2$$ $$|B|^2-2ARe\{B\times e^{+j\omega\delta}\}= |C|^2-2ARe\{C^*\times e^{+j\omega\delta}\}$$

given $B$ and $C$ are complex and $0<|A|\leq1 $.

Up to this point, I am stuck. I am not strong with math so I'm not sure how I can go about proving that $$B=C^*$$

Is it possible? Am I approaching the problem correctly? Any help will be much appreciated!

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

For any fixed complex numbers $w$ and $z$, the equation $$|X-w|=|Y-z|$$

has an infinite solution set $\{(X,Y)\}\subseteq \Bbb C^2$ that contains more than just pairs of complex conjugates.

Indeed, picking any $X\in\Bbb C\backslash\{w\}$ and defining $r=|X-w|$, the second component $Y$ will satisfy

$$|Y-z|=r,$$

which describes a circle of radius $r$ around $z$ in the complex plane. Thus there are infinitely many second components $Y$ associated to any $X\ne w$. Specialize to $w=Ae^{-j\omega\delta}$, $z=Ae^{+j\omega\delta}$.

share|improve this answer
    
Thanks anon! So even despite the fact that $w$ and $z$ are $w=z^*$ complex conjugate of each other. There are still infinite solutions to this problem? –  JuniorEngie May 2 '12 at 4:02
1  
@JuniorEngie: Not just an infinite number of solutions, an infinite number of solutions $X,Y$ that are not complex conjugates. –  anon May 2 '12 at 4:08
    
Ahh, cheers for the clarification. So basically this is an open-ended problem with no 'unique' solution. Meaning $X$ and $Y$ could be any complex value and the equation still hold true. Whilst meaning that $X=Y^*$ is "one" of the possible solutions. Hmm... then I need to find a way to prove that it is one of the possible solutions now! –  JuniorEngie May 2 '12 at 4:55
1  
@JuniorEngie: Prove that $|z|=|\bar{z}|$ for any $z\in\Bbb C$. Now notice that $\overline{B-Ae^{-j\omega\delta}}=\overline{B}-Ae^{+j\omega\delta}$. (I assume you're using $j$ as the imaginary unit, and $A,\omega,\delta$ are all real.) –  anon May 2 '12 at 4:57
1  
@Junior: You're welcome! –  anon May 2 '12 at 8:05
show 1 more 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.