Home ยป Inequality of complex numbers

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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