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Help with math proof

wannabe mathematician Thu 07 Jan 04:04:50 2021 No.12553476 View Reply Original Report

Quoted By: >>12553894

Anybody has any idea on how I could proof this.
I think I should probably use Rouche's Theorem but can't get any further.
Thanks

Anonymous

Anonymous Thu 07 Jan 2021 08:37:24 No.12553865 Report

Quoted By: >>12554583

Ask your professor like a normal human being.

Anonymous

Anonymous Thu 07 Jan 2021 08:38:13 No.12553868 Report

Quoted By:

Ask you prof.

Anonymous

Anonymous Thu 07 Jan 2021 09:05:30 No.12553894 Report

Quoted By: >>12554518 >>12554573 >>12554578

>>12553476

$a_{0}(t) = t$
$p(z; t) = z - a_{0}(t)$
$p(z; t)$ has $K = 1$ zeros at $t = 0$ in $D(a, r) = \textbf{B}(a, r) = \textbf{B}(-1, 1)$ , a closed ball at $a = -1 \in \mathbb{C}$ with radius $r = 1$ . When $t > 0$ , the root leaves the closed ball, so there aren’t $K$ zeros in $D(a, r)$ .

If you are going to prove that result, $D(a, r)$ must be an open ball, then the result follows trivially from the continuity of the $a_n(t)$ .

Anonymous

Anonymous Thu 07 Jan 2021 15:42:42 No.12554518 Report

Quoted By:

>>12553894
Yeah $D(a, r)$ is defined:
$D(a, r)=\{z \in \mathbb{C}:|z-a|<r\}$, but I can't see it's trivial. I have an idea but nothing concrete.

Anonymous

Anonymous Thu 07 Jan 2021 16:05:11 No.12554573 Report

Quoted By:

>>12553894
Yeah $D(a, r) [\math] is defined:<br /><script type="math/tex; mode=display">D(a, r)=\{z \in \mathbb{C}:|z-a|<r\}[\eqn].<br />I still can't see how it is trivial.$

Anonymous

Anonymous Thu 07 Jan 2021 16:06:50 No.12554578 Report

Quoted By:

>>12553894
Yeah $D(a, r)$ is defined:
$D(a, r)=\{z \in \mathbb{C}:|z-a|<r\}$ .
I still can't see how it is trivial.

Anonymous

Anonymous Thu 07 Jan 2021 16:08:29 No.12554583 Report

Quoted By:

>>12553865
If only my professor responded emails like a normal human being

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