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That’s the problem:

Let $f = x^7 - x^4 - 5x + 3i$ a complex polynomial. Prove that $f$ has at least a zero $\alpha \in \mathbb{C}$ with $| \alpha | < 1$.

The fact is that I’m supposed to solve this problem with topology and I have no idea about how to proceed. I would use Complex Analysis (Rouché Theorem) but that’s not the point. Is there anyone that can help me? Thank you.

Ulisse_Dini
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    Rouché's Theorem is topology. Winding numbers of curves around the origin are, indeed, topology. If you restrict to the unit circle, to what map is $f/|f|$ homotopic? – Ted Shifrin Dec 27 '22 at 20:35
  • It’s homotopic to the constant map. – Ulisse_Dini Dec 27 '22 at 21:42
  • How so? If so, why does that show $f$ has no roots inside the circle? – Ted Shifrin Dec 27 '22 at 22:12
  • I have no idea. I thought about the constant map thinking that it could be deformed in the polynomial I’m interested with. This part of the program for the exam has been explained in few hours so I have a lot of confusion in my head. – Ulisse_Dini Dec 28 '22 at 15:00
  • What is the exact instruction of the problem? Use topology, not use Rouche's theorem? Did you learn Rouche's theorem when you first learned about this problem? Anything specifically prohibited to use? – Sungjin Kim Dec 28 '22 at 20:31
  • It’s meant to be solved witho topology but not with Rouchè theorem because we haven’t seen it in this course. Nothing alse prohibited to use. – Ulisse_Dini Dec 29 '22 at 21:30
  • What material from topology did you learn up to this point when this problem is assigned? – Sungjin Kim Dec 30 '22 at 17:06
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    A simple approach with very minimal knowledge of algebraic topology: (1.) you want to show winding number $n\big(f\circ \gamma,0\big)\neq 0$ where $\gamma(t) = \exp\big(2\pi i \cdot t\big)$ for $t\in [0,1]$, as this implies a zero inside the unit circle for arbitrary $f$ that is continuous on the closed unit disc. (2.) show that $n\big(f\circ \gamma,0\big)$ exists-- i.e. that $f$ is non-zero on $S^1$. (3.) construct a linear homotopy from $f\circ \gamma$ to a $p\circ \gamma$ where $p$ is a monomial (part 2 should give you insights here) and the result follows by homotopy invariance – user8675309 Jan 01 '23 at 16:50
  • While choosing $p$ monomial would work, we may also use the comparison between $x^7-x^4 = x^4(x^3-1)$ and $-5x+3i$ on $S^1$. – Sungjin Kim Jan 03 '23 at 01:20

1 Answers1

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While Rouche's theorem and using winding numbers are the standard methods, I try to avoid them here as per the request.

We refer to Continuity of the roots of a polynomial in terms of its coefficients and a comment to the question in the link by Jyotirmoy Bhattacharya , who adresses this paper by Gary Harris and Clyde Martin: https://www.ams.org/journals/proc/1987-100-02/S0002-9939-1987-0884486-8/S0002-9939-1987-0884486-8.pdf, the proof there is elementary and topological.

The main point of the above post and the paper is that the roots of polynomials vary continuously with their coefficients.

The proof given in the paper assumes the polynomial is monic. The argument is adapted for non-monic polynomials with a fixed nonzero constant term. This can be done by considering the reciprocal polynomials.

We apply the above analysis to $P_t(z) = t(z^7-z^4)-5z+3i$. We let $t$ vary from $0\leq t \leq 1$. Then $P_0(z)=-5z+3i$ and $P_1(z)=f(z)$. We see that $P_0(z)$ has the unique root $z=3i/5$ inside $S^1$. Then the root $r_t$ of $P_t(z)$ with $r_0=3i/5$ forms a continuous path as $t$ varies from $0$ to $1$.

We prove that the path $r_t$ can never escape $S^1$.

For $z\in S^1$ and $0\leq t<1$, we have $$ |t(z^7-z^4)|\leq 2t < 2\leq |-5z+3i|. $$ For $z\in S^1$ and $t=1$, we have $$ |z^7-z^4|\leq 2\leq |-5z+3i|. $$ However, $2=|-5z+3i|$ occurs only with $z=i$ (use geometry of circles to prove this). On the other hand, $z=i$ does not satisfy $|z^7-z^4|=2$. Thus, we have $$ |z^7-z^4|<|-5z+3i|. $$ Hence, for $z\in S^1$, we must have $|z^7-z^4|<|-5z+3i|$. The path $r_t$ therefore does not cross $S^1$ when $t$ varies from $0$ to $1$.

Therefore, we proved that there exists at least a root of $f(z)$ with $|z|<1$.

Remark

Using the inequality $|z^7-z^4|<|-5z+3i|$ on $z\in S^1$, and apply Rouche's theorem, we can prove that we have exactly one root of $f(z)=0$ with $|z|<1$.

Sungjin Kim
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