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Let $n$ be a positive integer, $\lambda_{i}$ real numbers and $a_{i}$ for $1\leq i \leq n$ pairwise distinct complex numbers. Help me to prove that if $\forall z \in \mathbb{C}$ we have $ \sum(\lambda_{i}|z-a_{i}|)=0 $, then $\lambda_{i}=0$ for $ 1\leq i \leq n$.

Regards

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    @user10049 : It is a tricky one. I'll post a more detailed answer in a couple of minutes. (For now, a more detailed hint. If you want an answer, I'll give it, but you'll profit more from working through the hint.)2011-04-25

2 Answers 2

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If $\lambda_i \neq 0$, $|z-a_i| = -1/\lambda_i \sum_{j \neq i} \lambda_j |z-a_j|$, and the RHS is differentiable at $z=a_i$, while the LHS is not.

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    I like the solution a lot. Can you please enlighten me on how you came up with this solution?2011-04-25
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For now, let's fix $n = 2$. Then we have 2 real numbers $\lambda_1$ and $\lambda_2$, and two complex numbers $a_1$ and $a_2$. We want to see that if $\lambda_1 |z - a_1| + \lambda_2 |z - a_2| = 0$ (1) for all complex numbers $z$, then $\lambda_1 = \lambda_2 = 0$.

To do this, let's pick $z = a_1$ as a complex number. If equation (1) holds for all complex numbers, then it must hold for $z = a_1$. Thus we must have $\lambda_2 |a_1 - a_2| = 0$. As $a_1$ and $a_2$ are distinct by hypothesis, then this entails $\lambda_2 = 0$ (this would be a good place to construct a counterexample if the hypothesis fails). We now proceed either by recursion, or by picking another value for $z$, and prove that $\lambda_1 = 0$.

If you fill in the details and grok this argument, I think you'll have no trouble turning it into one that works for any given $n$.

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    FYI: induction on $n$ might be the way forward. To see why, try writing down the same argument for $n = 3$. See where you have to do extra work.2011-04-25