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Inequality of complex numbers (Solved)

Importance: Medium ✭✭
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Subject: Analysis
Keywords:
Recomm. for undergrads: yes
Posted by: feanor
on: April 7th, 2010
Solved by: fedorpetrov here in comments
Conjecture   There exists a real positive $ c $, such that for any $ n\in\mathbb{N} $ and any $ z_i\in\mathbb{C} $ where $ |z_i|\le 1 $ for $ 1\le i\le n $ and $ \~z:=\frac{1}{n}\sum^n_{k=1}z_k $, the following holds: $$\left|\prod^n_{k=1}z_k - \~z^n\right| \le c\cdot\sum^n_{k=1}|z_k-\~z|^2$$

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Yes, such exists, say

On January 5th, 2011 fedorpetrov says:

Yes, such $ c $ exists, say $ c=1000000 $ works. Assume the contrary and consider the counterexample. Without loss of generality, $ \max |z_i|=1 $, else multiple all $ z_i $'s to some $ \lambda>1 $ so that this bacomes true, LHS is multiplied by $ \lambda^n $, while RHS only by $ \lambda^2 $. So, we again get a counterexample. Denote $ \bar{z}=a $, $ z_i=a+y_i $, $ \sum y_i=0 $. Since LHS does not exceed 2, we have $ |y_i|<1/100 $ (else RHS is too large). Hence $ |a|=|z_i-y_i|>1-1/100 $ for $ i $ s.t. $ |z_i|=1 $. Then we have $ a+y_i=a(1+y_i/a)=ae^{y_i/a+p_i} $, where $ p_i=\ln(1+y_i/a)-y_i/a=(y_i/a)^2/2+(y_i/a)^3/3+\dots $, $ |p_i|\leq 2|y_i^2| $ by some easy estimate. Finally, LHS equals $$ |a^n|(e^{p_1+p_2+\dots+p_n}-1), $$ and we just use estimate $ |e^p-1|\leq 2p $ for small enough $ p=p_1+p_2+\dots+p_n $ ($ p $ is small enough, since $ |p|\leq 2\sum |y_i^2|=\frac2{c}RHS\leq \frac4{c} $).

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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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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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Requesting background information

On April 21st, 2010 Carolus says:

To feanor, the author of this conjecture:

What is the motivation for this conjecture ? The selected importance "medium" let me assume the verification or falsification of this conjecture would bring some benefit. If possible, describe that benefit ("practical" applications or consequences), please.

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