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 $p_n(a)\neq 0$
For the case $|a|=1$, I think i have a partial solution: Suppose that $a\in \mathbb{C}$ is such that $p_n(a)=0$ and $|a|=1$, then by the revers triangle inequality $$ 0 = | p_n(a)| > \left| 1 - \left|a+\frac{a^2}{2}+\cdots+\frac{a^n}{n!}\right|\right| $$ which is a contradiction, because since $n\geq 2$, the RHS of the last inequality must be positive (I said partial solution because I am not sure how to prove this, I am almost sure that this follows because $| a + \cdots a^n/n!|>1$ if $n\geq 2$ but I can't prove that either) EDIT: According to the comments this is wrong, so my question now extends to both cases!.
For the case $|a|=n$, i tried something similar but it gets worst.
My questions are: 1) Is my approach for the first case correct? If it is, how can I prove the details I am missing, and if is not how can I approach it? 2) How can I approach the second case ? Any help or hints will be very appreciated
Bringing down the solution in the comments given by User Kevin Arlin in order to take this question off the unanswered list.
Note that the given polynomial $p_n(z)$ is the $n$th order Taylor polynomial of $f(z) = e^z$. Suppose $\lvert a \rvert = 1$. On the one hand, we have $1/e \leq \lvert e^a \rvert \leq e$. On the other hand, $$ \lvert e^a - p_n(a) \rvert = \left\lvert \sum_{m = n+1}^\infty \frac{a^m}{m!} \right\rvert \leq \sum_{m = n+1}^\infty \frac{1}{m!}. $$ So, if $a$ were a zero of $p_n(z)$, then we would have $$ \frac{1}{e} \leq \lvert e^a \rvert = \lvert e^a - p_n(a) \rvert \leq \sum_{m = n+1}^\infty \frac{1}{m!}. $$ But, for $n \geq 2$, we have $$ \sum_{m = n+1}^\infty \frac{1}{m!} \leq e - \frac{1}{0!} - \frac{1}{1!} - \frac{1}{2!} = e - 5/2. $$ However, $$ \frac{1}{e} > e - \frac{5}{2}, $$ a contradiction.
Hence, if $a$ is a zero of $p_n(z)$ for $n \geq 2$, then $\lvert a \rvert \neq 1$.
(It might be worth noting that at $n = 1$, we have $p_1(z) = 1 + z$, which does have a zero of unit modulus, namely $a = -1$. When $n = 0$, we have $p_0(z) = 1$, which has no zeros at all, but this is a trivial scenario.)
The case $\lvert a \rvert = n$ is more complicated, and is dealt with in this question: Show that $p_n(a)\neq 0$ if $|a|=n$.