# A Romanian mathematical olympiad problem

Let $z_{1}$, $z_{2}$, $\ldots$, $z_{n}$ be complex numbers such that $|z_{1}|=|z_{2}|=|z_{3}|= \ldots = |z_{n}|>0$. Prove that

$\Re (\Sigma_{j=1}^{n}\Sigma_{k=1}^{n}\frac{z_{j}}{z_{k}})=0$

if and only if $\Sigma_{k=1}^{n}z_{k}=0$

(Romanian Mathematical Olympiad — Second Round, 1987)

Proof:

Let $S=\Sigma_{j=1}^{n}\Sigma_{k=1}^{n}\frac{z_{j}}{z_{k}}$.

Then, $S=(\Sigma_{k=1}^{n}z_{k}).(\Sigma_{k=1}^{n}\frac{1}{z_{k}})$ and since $z_{k}.\overline{z_{k}}=r^{2}$ for all k, we have

$S=(\Sigma_{k=1}^{n}z_{k}).(\Sigma_{k=1}^{n}\frac{\overline{z_{k}}}{r^{2}})$

which equals $\frac{1}{r^{2}}(\Sigma_{k=1}^{n}z_{k}).(\overline{\Sigma_{k=1}^{n}z_{k}})=\frac{1}{r^{2}}|\Sigma_{k=1}^{n}z_{k}|^{2}$

Hence, S is a real number, so $\Re{S}=S=0$, if and only if $\Sigma_{k=1}^{n}z_{k}=0$.

More later,

Nalin Pithwa

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