Home ยป Inequality of complex numbers

A hint: \par Let with

On July 29th, 2010 Anonymous says:

A hint: \par Let $ z_k=\bar z+\epsilon a_k $ with $ \epsilon $ small and $ \sum_k a_k=0 $. Then $$\product_k z_k - \bar z^n = \epsilon^2 \bar z^{n-2} \sum_{j<k} a_j a_k + ?\epsilon^3$$ and $ |z_k - \bar z|^2 = \epsilon^2 |a_k|^2 $. \par Now from $ \sum_k a_k=0 $ it follows that $$| \sum_{j<k} a_j a_k | \le {1\over 2} \sum_k |a_k|^2 .$$ This shows that your inequality has ``the right order of magnitude'' with $ c={1\over 2} $.

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