Let $a\in\mathbb{C}$, and let $n \ge 2$. Prove that the polynomial $2022+az+2023z^n$ has a root in the unit disk, $D(0,1)$.
There's an algebraic way to solve this with Vieta's formulas, by observing that the product of all of the roots is $(-1)^n\cdot 2022/2023$, which has absolute value less than $1$, and so there must be a root with modulus less than $1$.
I'm wondering if there's a way to solve this using analysis. Specifically, I'm wondering about using Rouché's theorem.
Breaking it down into cases, in the case $|a|>2022+2023$, then for all $z \in \partial D(0,1)$, $|2023z^n+2022|\le2023|z|^n+2022=2023+2022<|a|=|az|$. Thus, by Rouché's theorem, since $az$ has a root in $D(0,1)$, $2023z^n+az+2022$ has a root in $D(0,1)$ as well.
Since $a$ can vary, it seems right to focus our attention on $a$. However, I can't get past the case that $|a|<2022+2023$.
This question is from a complex analysis qual at my uni, which is why I prefer to find an analysis-based way to solve this problem, rather than an algebraic. It's certainly possible that there's an analysis based way to approach this that doesn't entail Rouché's theorem, though the problem statement heavily primes us to use Rouché.
Best Answer
Here's a proof based on Schur transformations and in particular ex 4.2.26 in Ash & Novinger's Complex Variables, an exercise which I think OP should carefully go through so the underlying ideas solidify.
(a) $p(z)=c_n z^n + \dots + c_1 z + c_0=2023z^n+az+2022$
(b) $q(z) := \overline c_0 z^n + \dots + \overline c_{n-1}z + \overline c_n=2022z^n + \overline az^{n-1} + 2023$
$ =z^n\cdot \overline{p\Big( \big(\overline z\big)^{-1}\Big)}$ for $z\neq 0$
(c) Schur Transform $f(z) := \overline {c_0}\cdot p(z) - c_n\cdot q(z)=2022\cdot p(z) - 2023\cdot q(z)$ noticing in our case $\text{degree } f = n-1$
Initially we assume that $0\notin p\big(S^1\big)$. The three relevant properties of $q$ are
(i.) $\big \vert p(z)\big \vert = \big \vert q(z)\big \vert$ for $z\in S^1$ [write $q(z)=z^n\cdot \overline {p(z)}$ for $z\in S^1$]
(ii.) if $p$ has $k$ zeros in $B\big(0,1\big)$ then $q$ has $n-k$ zeros in $B\big(0,1\big)$ [use the Argument Principle]
(iii.) $\vert c_n\vert \gt \vert c_0\vert\implies f \text{ has } n-k$ zeros in the open disc. This is an immediate application of Rouche:
$\vert \overline c_0 p(z)\vert \lt \big\vert -c_n q(z)\big\vert$ for $z\in S^1$ so $f$ and $-c_n q(z)$ have the same number of roots in $D$, i.e. $n-k$. Now $\text{degree }f =n-1\implies k\geq 1$, so $p(z)$ has a root in the (open) unit disc.
nuisance case: zeros in the unit circle
Let $p$ have $r$ zeros (with multiplicity) in $S^1$. Then $p(z)=P(z)\cdot\prod_{j=1}^r(z-\lambda_j)$ where $0\notin P(S^1)$. Property (i.) tells us that $q(z)=Q(z)\cdot\prod_{j=1}^r(z-\lambda_j)$ where $0\notin Q(S^1)$ and $\vert Q(z)\vert=\vert P(z)\vert$ i.e. (i.) still holds since $S^1$ is connected, so its image under $\big \vert P(z)\big \vert -\big \vert Q(z)\big \vert$ is connected. Observe $f(z) = F(z)\cdot\prod_{j=1}^r(z-\lambda_j)$ and $r\lt n$ since $\text{degree }f =n-1$. Repeating the earlier Rouche argument $F(z)$ and $Q(z)$ have the same number of zeros in the unit disc. We argue below that if $P(z)$ (hence $p(z)$) has $k$ zeros in $B\big(0,1\big)$, then $Q(z)$ has $(n-r)-k$ zeros in $B\big(0,1\big)$ and as before $k\geq 1$ because $F(z)$ (hence $f(z)$) cannot have $n-r$ zeros in $B\big(0,1\big)$ since $\text{degree }f = n-1$.
$p_m(z):= P(z)\cdot\prod_{j=1}^r(z-\frac{m+1}{m}\lambda_j)$. Then $p_m(z)$ has $k$ zeros in the open unit disc and none on its boundary for all $m$ and hence its reciprocal polynomial $q_m(z)$ (which varies continuously with $p_m$) has $n-k$ zeros in the (open) unit disc for all $m\in \mathbb N$ per the preceding (non-nuisance) argument. Define $\gamma, \gamma_\delta:[0,1]\longrightarrow \mathbb C$ given by $\gamma(t)=\exp\big(2\pi i\cdot t\big)$ and $ \gamma_\delta(t) =(1+\delta)\cdot\exp\big(2\pi i\cdot t\big)$ for any $\delta \gt 0$ small enough-- in particular via uniform continuity of $Q(z)$ on $\overline B\big(0,2\big)$ choose $\delta$ such that $Q(z)$ is non-zero on the closed annulus $1-\delta\leq \vert z \vert\leq 1+\delta$. Finish by applying the Argument Principle and Hurwitz as $q_m\to q$:
$n-k= n\big(q_m\circ \gamma_\delta,0\big)= n\big(q\circ \gamma_\delta,0\big)=\big(Q\circ \gamma_\delta,0\big)+ r=\big(Q\circ \gamma,0\big)+ r$
i.e. $n\big(Q\circ \gamma,0\big) = (n-r)-k$ which means degree $n-r$ polynomial $Q(z)$ has $(n-r)-k$ zeros in the unit disc per the Argument Principle.