Linear Independence of Complex Absolute Value Function

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.

• 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. – phdmba7of12 Oct 4 '18 at 17:37
• intuition tells me the equation ... always represents a curve Indeed, that's a generalized conic of the form of a multifocal oval curve. – dxiv Oct 6 '18 at 2:39