Given $n$ different complex numbers $z_1,z_2\cdots z_n$, $n$ real numbers $a_1,a_2\cdots a_n$ and a constant $C$, if $$\sum_{i=1}^{n}a_i\left|z-z_i\right|=C$$ for every complex number $z$ in a region $\Omega$ on the complex plane, will it imply that $a_i=C=0\quad\left(i=1,2,\cdots n\right)$?
i.e. functions $\left|z-z_i\right|$ are linearly independent in the space $\mathbb{R}^\Omega$?
If not, does there exist a counterexample?
I think it’s unlikely to be false, because intuition tells me the equation $\sum_{i=1}^{n}a_i\left|z-z_i\right|=C$ (if non-trivial) always represents a curve in $\Omega$.

I noticed recently that it can be converted into the following question: If $$\sum_{i=1}^n a_i\frac{\left(z-z_i\right)}{\left|z-z_i\right|}=0$$for all $z\in \Omega$, prove $a_i=0,\quad i=1,2\cdots n$.
My thought is to pick $z=Z_1,Z_2\cdots Z_n$ such that the matrix $$\left(\frac{\Re\left(Z_i-z_j\right)}{\left|Z_i-z_j\right|}\right)_{n\times n}\quad or \quad\left(\frac{\Im\left(Z_i-z_j\right)}{\left|Z_i-z_j\right|}\right)_{n\times n} $$ has nonzero determinant.

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    $\begingroup$ if the sum of distances between a point $z$ and a set of points $\left{z_i\right}$ is the same for all $z$ in a region $\Omega$, then intuition suggests that if $n=2$ this region $\Omega$ is a straight line and only two of $a_i$ are non zero. $\endgroup$ – phdmba7of12 Oct 4 '18 at 17:37
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    $\begingroup$ intuition tells me the equation ... always represents a curve Indeed, that's a generalized conic of the form of a multifocal oval curve. $\endgroup$ – dxiv Oct 6 '18 at 2:39

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