I am working the next problem:
Consider the polynomials $$ p_n(z)=\sum_{j=0}^{n}\frac{z^j}{j!} $$ For $n \geq 2$, show that if $a \in \mathbb{C}$ is such that $|a|=1$ or $|a|=n$, then $p_n(a)\neq 0$
The case $|a|=1$ has already been answer here on the last comment of this question, however I still have no clue for the case $|a|=n$. Any help? I have tried a lot of things, as you can see on the other question but nothing good.
Looking back, it appears that I was barking up the wrong tree with Rouche's theorem.
It suffices to show that the polynomials $q_{n}(z):=p_{n}(nz)$ have no zeros on the unit circle for $n\geq 2$. We can and will, in fact, prove a stronger result: $q_{n}$ has no zeros outside the open unit disk $D(0,1)$.
Lemma 1. Let $p(z)=a_{0}+a_{1}z+\cdots+a_{n}z^{n}$ be a polynomial of degree $n>2$ with nondecreasing positive coefficients $0<a_{0}\leq a_{1}\leq\cdots\leq a_{n}$.
Proof. Consider the polynomial $f(z):=(1-z)p(z)$. The zeros of $f$ are precisely the zeros of $p$ together with $z=1$. So if we show that $f$ has no zeros outside the open unit disk except $z=1$, then we're done. Observe that
$$f(z)=a_{0}-a_{n}z^{n+1}+\sum_{j=1}^{n}\left(a_{j}-a_{j-1}\right)z^{j}\tag{1}$$
Suppose that $f(z)=0$ for $\left|z\right|\geq 1$. Then
\begin{align*} \left|a_{n}z^{n}\right|&=\left|a_{0}+\sum_{j=1}^{n}\left(a_{j}-a_{j-1}\right)z^{j}\right|\\ \\ &\leq a_{0}+\sum_{j=1}^{n}\left|\left(a_{j}-a_{j-1}\right)z^{j}\right|\tag{2}\\ \\ &\leq a_{0}+\sum_{j=1}^{n}\left(a_{j}-a_{j-1}\right)\left|z\right|^{n}\tag{3}\\ \\&=a_{n}\left|z\right|^{n} \end{align*} If $\left|z\right|>1$, then the inequality in (3) is strict and we obtain a contradiction, which completes the proof of the first claim.
Now suppose the hypotheses of the second claim hold. If $\left|z\right|=1$, then the inequality in (2) is in fact an equality. Whence by Lemma 2 (see below), $(a_{j}-a_{j-1})z^{j}=t(a_{j'}-a_{j'-1})z^{j'}$ for some real $t>0$. Without loss of generality, assume $j'=j+1$ so that
$$z=\dfrac{a_{j}-a_{j-1}}{t(a_{j+1}-a_{j})}\in\mathbb{R}^{>0}$$
Since $\left|z\right|=1$, we conclude that $z=1$. But clearly $p(1)\neq 0$. $\Box$
You can check that for $n\geq 2$ the coefficients of
$$q_{n}(z)=\sum_{j=0}^{n}\dfrac{n^{j}}{j!}z^{j}$$
satisfy both hypotheses of Lemma 1. They're all strictly increasing except $n^{n}/n!=n^{n-1}/(n-1)!$. When $n=2$, you can use the quadratic formula to directly compute that $p_{2}(z)=0\Leftrightarrow z=-1\pm i$.
Lemma 2. Suppose $z_{1},\ldots,z_{m}\in\mathbb{C}$ are nonzero, and set $z:=z_{1}+\cdots+z_{m}$. If
$$\left|z\right|=\left|z_{1}\right|+\cdots+\left|z_{m}\right|,$$
then $z_{j}=\lambda_{j}z$, where the $0\leq\lambda_{j}\leq 1$ and $\sum_{j}\lambda_{j}=1$.
Proof. The proof is basically induction on $m$ together with recalling that (as a consequence of Cauchy-Schwarz) the triangle inequality $\left|z+w\right|\leq\left|z\right|+\left|w\right|$ is strict unless $z=t w$, for some real $t\geq 0$.
Suppose the result is true for $m-1$ vectors. The hypothesis implies that
$$\left|z\right|=\left|z_{1}\right|+\left|z_{2}+\cdots+z_{m}\right|,$$
whence $z_{1}=\lambda(z_{2}+\cdots+z_{m})$, for some real $\lambda>0$. Therefore
$$z_{1}=\dfrac{\lambda}{1+\lambda}z,\quad z_{2}+\cdots+z_{m}=\dfrac{1}{1+\lambda}z$$
We now apply the induction hypothesis to $\left|z_{2}+\cdots+z_{m}\right|=\left|z_{2}\right|+\cdots+\left|z_{m}\right|$ to obtain that
$$z_{j}=\tilde{\lambda}_{j}(z_{2}+\cdots+z_{m})=\underbrace{\dfrac{\tilde{\lambda}_{j}}{1+\lambda}}_{:=\lambda_{j}}z, \qquad 2\leq j\leq m$$
where $0<\tilde{\lambda}_{j}<1$ and $\sum_{j}\tilde{\lambda}_{j}=1$. Setting $\lambda_{1}:=\lambda/(1+\lambda)$ completes the proof of the induction step. $\Box$
This can be shown using a sharpened Kakeya-Enestrom theorem due to Anderson, Saff, and Varga. The paper (unfortunately missing pages 12 and 13) is available from Saff's webpage: PDF link.
In particular, in the notation of that paper $\beta = n$, so by Corollary 2 all zeros of $p_n(z)$ satisfy $|z| < n$.
We note that the same result can be applied to the reciprocal polynomial $q_n(z) = z^n p_n(1/z)$. In that case it says that all zeros of $q_n(z)$ lie in $|z|<1$ and hence all zeros of $p_n(z)$ lie in $|z|>1$.
