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 $n$.