# Roots of a polynomial

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

• For the first part, I'd use that $p_n$ is a Taylor polynomial for $e^x$. If $|a|=1$ then $|e^a|\in [1/e,e]$ and $|p_n(a)-e^a|=|\sum_{m>n} a^m/m!|\leq \sum_{m>n} 1/m!$. The worst case is $n=2$, when we're considering $1/3!+1/4!+...=e-5/2<1/e$. Commented May 31, 2015 at 0:52
• @KevinCarlson Thanks ! I do not quite see how your comment gives that $p(a)$ can not be zero, but I'll follow your idea to see if I can find out. On the other side what do you think of my argument for the first part ? And how about the second one, any thoughts ? Thanks again Commented May 31, 2015 at 1:51
• @LeoSera The part in your partial solution that you could not prove is not true. For example fix $n=3$, if $a=e^{\pi i}$, clearly $|a|=1$, however note that $| a + a^2/2 + a^3/6|=2/3 < 1$. Nevertheless I think you are on the right track, since you already got that $0>|1-|\cdots| | \geq 0$ a contradiction! For your second case I still have no clue, if I get something I´ll post it as an answer. Commented May 31, 2015 at 2:15
• @LeoSera Well, you don't have an argument for the first part yet. It might work, or it might not. I'm not sure how to do the second part. Commented May 31, 2015 at 20:32
• @KevinCarlson Thanks again, I see from AlonsoDelfín comment that my first part is wrong :(. However I still not see from your first comment how it follows that $p(a)\neq 0$, could you elaborate? Commented May 31, 2015 at 23:33

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 $|a|=n$.