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Infer from the theorem of Rouché that every non-constant polynomial does have a zero point in $\mathbb{C}$ (Fundamental Theorem of Algebra).

Consider the polynomial $$ p(z)=\sum_{i=0}^{n}a_iz^i, a_i,z\in\mathbb{C}, a_i\neq 0. $$ In order tu use Rouché, set $$ f(z):=a_nz^n,~~~~~~~~~~g(z):=\sum_{i=0}^{n-1}a_i z^i. $$ Because of the holomorphism of polyniomials, these functions are holomorphic and for each $r>0$ it is $$ f(z)\neq 0~\forall z\in\mbox{rg}(\gamma_r),\\ \gamma_r\colon [0,2\pi]\to\mathbb{C}, t\longmapsto r\exp(it). $$

Because of $$ \frac{\lvert z\rvert^i}{\lvert z\rvert^n}\to 0\mbox{ for }\lvert z\rvert\to\infty, i<n $$ it follows that $$ \lim\limits_{\lvert z\rvert\to\infty}\frac{\lvert g(z)\rvert}{\lvert f(z)\rvert}\leq\lim\limits_{\lvert z\rvert\to\infty}\frac{\sum_{i=0}^{n-1}\lvert a_i\rvert\cdot\lvert z^i\rvert}{\lvert a_n\rvert\cdot\lvert z^n\rvert}=\frac{1}{\lvert a_n\rvert}\lim\limits_{\lvert z\rvert\to\infty}\sum_{i=0}^{n-1}\lvert a_i\rvert\cdot\lvert z\rvert^{i-n}=0. $$ So there exists an $\varepsilon>0$, so that $$ \frac{\lvert g(z)\rvert}{\lvert f(z)\rvert}<1~\forall z\in\mbox{rg}(\gamma_{\varepsilon}). $$ So let $\gamma_{\varepsilon}$ be the curve tracing the open circular disk with radius $\varepsilon$. The theorem of Rouché then says that the number of zero points of $p$ within that circular disk is the same as the number of zero points of $f$ within that circular disk. And within that disk, it is $f(z)=0$ for $z=0$ (and nowhere else). So $p$ indeed has a zero point in $\mathbb{C}$.

Do you agree? Is my proof okay?

tejasvi88
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    Yes, the proof is okay. You may ponder including that the multiplicity of $f$'s root is $n$, hence $p$ has $n$ zeros (counting multiplicity) in the disk. – Daniel Fischer Sep 08 '13 at 13:04
  • Thank you! $0$ is a zero point of multiplicity $n$n. So $p$ has n zero points (all the same, i.e. 0). –  Sep 08 '13 at 18:24
  • $f$ has one root with multiplicity $n$. $p$ has most likely $n$ distinct roots. – Daniel Fischer Sep 08 '13 at 18:26
  • Do not know exactly what you mean. May you explain it? –  Sep 08 '13 at 18:49
  • In your comment, you wrote "So $p$ has $n$ zero points (all the same, i.e. $0$)." But that is true of $f = a_n z^n$, not - except in the trivial case - of $p$. I guess you just mixed $f$ and $p$ up in your comment. $p$ has in general $n$ distinct roots. – Daniel Fischer Sep 08 '13 at 18:53
  • Okay, now I see! But I think the main point is that $p$ has n roots, because f has n roots. What multiplicity the roots of p have is not important I guess. –  Sep 08 '13 at 19:03
  • Right, that's the main point. You don't know what multiplicities the roots of $p$ have, but you know the sum of the multiplicities. Then there's the general observation that "almost all" polynomials have no multiple roots, so it's likely that $p$ has $n$ distinct roots. – Daniel Fischer Sep 08 '13 at 19:06
  • Thanks a lot! You live and learn. :-) –  Sep 08 '13 at 19:08

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