How can I show that $z^4-2z+3$, $z \in \mathbb{C}$, has no zeros within the unit circle in the complex plane? It looks like the Rouche theorem, but i still cannot do it. Please help. Thanks in advance.
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Thanks, that helps. :) – Logic_Problem_42 Feb 07 '22 at 14:00
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But we dont have |f(z)|>|g(z)| in all z with |z|=1. Do we? – Logic_Problem_42 Feb 07 '22 at 14:03
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Yes, I figured it out already. Thanks. – Logic_Problem_42 Feb 07 '22 at 14:15
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You do not need Rouche, just the triangle inequality. See the answer. – Ryszard Szwarc Feb 07 '22 at 14:24
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Edited because $4z^4-2z+3$ does have zeroes inside the unit circle. – JMP Feb 07 '22 at 14:57
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You are right. In this case the product of all the roots is equal ${3\over 4}.$ – Ryszard Szwarc Feb 07 '22 at 15:22
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1@JeanMarie; its from a previous edit by accident I presume – JMP Feb 07 '22 at 17:24
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We have $$|z^4-2z+3|\ge 3-2|z|-|z|^4.$$ Hence the equation does not admit solutions for $|z|<1.$ In the case $|z|=1$ the equality $z^4-2z=-3$ implies $z^4=-1$ and $z=1,$ which leads to a contradiction.
Ryszard Szwarc
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@RobArthan $|a+b+c|\ge |c|-|a|-|b|,$ where $a=z^4,$ $b=-2z$ and $c=3.$ – Ryszard Szwarc Apr 01 '22 at 01:01
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And how does that inequality hold? I have looked for a proof but didn't find any.Thanks. – CharlesJA Sep 14 '23 at 17:06
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A proof using Rouché's theorem.
Consider $f(z):=z^4+3$, $g(z):=−2z$.
On the disk $D_{\varepsilon}$ with center 0 and radius $1−\varepsilon$. On $\partial D_{\varepsilon}$,
$$|f(z)| \ge 2+3 \varepsilon \ \ \text{and} \ \ |g(z)| \le 2(1−\varepsilon)$$
for $\varepsilon$ sufficiently small, allowing Rouché's theorem to be applied, proving that there is no zero inside $D_{\varepsilon}$.
(thanks to Lutz Lehmann who has pointed an error of mine, now fixed).
Jean Marie
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Ok, yes, that bothersome to have missed the signs. Still, how, or better why, do get to $2+ε^4$? For $ε<\frac12$ one gets $...>2+ε+3ε^3>2+ε$ – Lutz Lehmann Mar 31 '22 at 18:32
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In fact, You are right, I should have written $2+3 \varepsilon + ( \varepsilon- 6\varepsilon^2 + 4 \varepsilon^3 - \varepsilon^4) > 2+3 \varepsilon$ for $\varepsilon>0$ sufficiently small. I will correct it – Jean Marie Mar 31 '22 at 18:58