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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