Home ยป Inequality of complex numbers

A hint

On August 1st, 2010 Christian Blatter says:

I am the author of this hint and somehow mismanaged the posting. So here it is again:

Let $ z_k=\bar z + \epsilon a_k $ with $ \epsilon $ small and $ \sum_k a_k=0 $. Then $$\prod_k z_k - \bar z^n = \epsilon^2 \bar z^{n-2}\sum_{j<k}a_j a_k + ?\epsilon^3$$ and $ |z_k-\bar z|^2=\epsilon^2|a_k|^2 $.

Now from $ \sum_k a_k=0 $ it follows that $$|\sum_{j<k} a_j a_k| \le {1\over2} \sum_k|a_k|^2.$$ This shows that your inequality has the "right order of magnitude" with $ c={1\over2} $.

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