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For a complex polynomial $\sum_0^n a_i z^i$ the sum of roots is readily given by the Vieta's formula. However, after trying to find out the sum of moduli of roots in terms of the coefficients, I feel like it should not be algebraically computable in terms of coefficients. Then I tried to figure out some bounds for the mean of moduli. I realised that we can trivially compute a class of bounds by replacing each modulus by some bound for roots like Cauchy's bound. However, I could not think of some better way. I have the following questions:
Q.No1 Is it possible to find some good bounds for the average of the moduli of roots of a complex polynomial with complex roots with known coefficients?
Q.No.2 Now consider a random polynomial of degree $n$ with coefficients distributed as iid standard normal variates. How can we find the expectation of the mean of moduli of the roots, especially for finite degree polynomials?
Of the many things I was trying, maybe the following is of some interest. For any polynomial $\sum_0^n a_i z^i,$ we may note that $$\text{ |Product of roots| }=\left|\frac{a_0}{a_n} \right|$$ so if $z_{min}$ is the root min minimum modulus, then $$ |z_{min}| \geq \left(\left|\frac{a_0}{a_n} \right| \right)^{1/n} $$ so that $$ \frac{\sum_1^n |z_i|}{n} \geq \left(\left|\frac{a_0}{a_n} \right| \right)^{1/n} $$ I don't know, however, how good that bound is and whether we can find some good ones. I will be obliged for any help/links/suggestions

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  • $\begingroup$ May be related to your question: mathoverflow.net/questions/46068/… $\endgroup$ Commented Mar 10, 2023 at 16:32
  • $\begingroup$ Can you solve the problem for real quadratics, Agnost? $\endgroup$ Commented Mar 10, 2023 at 21:45
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    $\begingroup$ For a quadratic $ax^2+bx+c $ if the roots are $z_1,z_2$,then sum of squares of deviations is $$ | z_1-(z_1+z_2)/2|^2+| z_2-(z_1+z_2)/2|^2$$ ,which simplifies to $$ \frac{|z_1-z_2|^2}{2}.$$ Using $|x-y|^2=|(x+y)^2-4xy|$,and Vieta's formulae, we get $\sigma^2=\frac{|D|}{4|a|^2}$,where $D $ is the discriminant $\endgroup$ Commented Mar 11, 2023 at 4:12
  • $\begingroup$ If you want to ensure that I see a comment intended for me, you have to include @Gerry in it. $\endgroup$ Commented Mar 15, 2023 at 12:42

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