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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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    @Junior: You're welcome!2012-05-02