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Let $n>1$ and $c_0>c_1>c_2>\dots >c_n>0$ and $f(z)=c_0+c_1 z+\dots + c_n z^n$. Show that this polynomial doesnt have zeros in a unit ball $B(0,1)$.

Can you give me some feedback?

luka5z
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    If it's of any help, you can suppose without loss of generality that $c_0 = 1$. I realized this by considering the limit case $c_0 = \cdots = c_n = 1$, in which case the roots are $n+1^{\text{th}}$ roots of unity. I am still thinking about the problem though. – Patrick Da Silva Oct 19 '14 at 13:15
  • An idea which I don't know works is to bound the integral of $f'/f$. If the absolute value (divided by $2\pi i$) is less than 1 along the unit circle, it must be zero. – Aaron Oct 19 '14 at 13:21
  • my professor told us that this can be done without refering to integrals – luka5z Oct 19 '14 at 13:27
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    (Take $1/z$ instead of $z$ to see that this question is a duplicate of the one I linked to.) – Hans Lundmark Oct 19 '14 at 13:28

1 Answers1

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Consider $$(z-1)f(z)=-c_0+(c_0-c_1)z+(c_1-c_2)z^2+\ldots+c_nz^{n+1}.$$ If $|z|<1$ then by the triangle inequality $$|(c_0-c_1)z+(c_1-c_2)z^2+\ldots+c_nz^{n+1}|<c_0-c_1+c_1-c_2+\ldots+c_n=c_0$$ and so $(z-1)f(z)\neq 0$.

WimC
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