# Finding roots of complex polynomial

Consider the polynomial of degree $n$, we have $$P_n(z)=z+\frac{z^2}{2}+...+\frac{z^n}{n!}$$.

Arrange all of its roots $(z_{n,1},z_{n,2},...,z_{n,n})$ in increasing order of magnitude $$0=|z_{n,1}|\leq |z_{n,2}|\leq ... \leq |z_{n,n}|.$$

Show that $\lim\limits _{n\to\infty}|z_{n,2}|=2\pi$.

Thoughts so far:

Very generally, I would like to show that $|z_{n,2}|$ < $|z_{n,3}|$ so that we can have r>0 so that the disc |z|< r containing only 2 roots of $P_{n}$: $z_{n,2}$ and 0. Then I would like to integrate $\frac{1}{P_{n}}$ over |z| = r, and use the residue theorem to get an expression in terms of $z_{n,2}$, and then solve for $z_{n,2}$. Then, I'd take the limit as n $\rightarrow \inft However, I'm not having a lot of success with this method... any suggestions for another way to approach the problem? Thanks! - add comment ## 2 Answers IMHO, it is sufficient to show that the sequence of polynomials$\{P_n(z)\}_{n\in\mathbb{N}_0}$is uniformly convergent to the function$e^z-1$on any compact set$\mathbb{D}_r=\{z\in\mathbb{C}:|z|\leq r\}$. The zeros of$e^z-1$are simple and isolated, so, under uniform convergence, the zeros of$P_n(z)$must converge towards the well-known zeros of$e^z-1$. Uniform convergence is quite easy to show, since, on$\mathbb{D}_r$, $$\|(e^z-1)-P_n(z)\|_{\infty}\leq\frac{r^n}{n!},$$ and the RHS tends to zero when$n$approaches$+\infty$. - add comment HINT Note that$\displaystyle \lim_{n \to \infty} P_n(z) = \exp(z) - 1$. The roots of$\exp(z) - 1 = 0$are given by$z = 2 k\pi i$, where$k \in \mathbb{Z}$. Since$P_n(z) \to \exp(z)-1$,$z_{n,2} \to \pm 2 \pi i$. (Why?) - @ABCBach Yes, Almost. Note that$\lim_{n \to \infty} z_{n,2} = 2 \pi i$,$\lim_{n \to \infty} z_{n,3} = -2 \pi i$,$\lim_{n \to \infty} z_{n,4} = 4 \pi i$,$\lim_{n \to \infty} z_{n,5} = -4 \pi i$and in general$\lim_{n \to \infty} z_{n,k} = (-1)^k 2 \lfloor k/2 \rfloor \pi i\$ –  Please delete account Nov 18 '12 at 4:48
If the function converges, how do we know that the roots necessarily converge? –  ABC Bach Nov 19 '12 at 21:02