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

  • $\begingroup$ 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 $1/3!+1/4!+...=e-5/2<1/e$. $\endgroup$ Commented May 31, 2015 at 0:52
  • $\begingroup$ @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 $\endgroup$
    – Leo Sera
    Commented May 31, 2015 at 1:51
  • 2
    $\begingroup$ @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. $\endgroup$ Commented May 31, 2015 at 2:15
  • $\begingroup$ @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. $\endgroup$ Commented May 31, 2015 at 20:32
  • $\begingroup$ @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? $\endgroup$
    – Leo Sera
    Commented May 31, 2015 at 23:33

1 Answer 1


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 $|a|=n$.


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