Can we proof $B=C^*$ given $|B-Ae^{-j\omega\delta}|=|C-Ae^{+j\omega\delta}|$

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!

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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 $w=Ae^{-j\omega\delta}$, $z=Ae^{+j\omega\delta}$.

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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
@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
@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
@Junior: You're welcome! –  anon May 2 '12 at 8:05