Zeros of polynomials via Topology 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.
 A: 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 $|z|<1$.
