Complex zeros of the polynomials $\sum_{k=0}^{n} z^k/k!$, inside balls

this is a question from a Temple prelim exam, and i'm trapped in it! We have $p_n(z)=\sum_{k=0}^n\frac{z^k}{k!}$ and we have to prove that $\forall r>0 \quad \exists N\in\mathbb{N}$ s.t. $p_n(z)$ has no zeros in $B_r(0)$ for $n>N$.

I tried to use Rouche's theorem observing that $1+\frac{z^n}{n!}$ has the nth roots of $n!$ as complex zeros, which are outside a fixed ball for n sufficiently large, but i'm having trouble proving that $|1+\frac{z^n}{n!}|>|z+\cdots+\frac{z^{n-1}}{(n-1)!}|$ in $|z|=r$ (if this is true). If anyone could help..

It was proved by K. S. K. Iyengar in 1938 [1] that every zero of the polynomial

$$s_n(z) = \sum_{k=0}^{n} \frac{z^k}{k!}$$

lies in the annulus

$$\frac{n}{e^2} &lt |z| &lt n.$$

A much stronger result was proved in 1966 by Buckholtz [2], namely that the rescaled polynomials $s_n(n z)$ have no zeros in the region

$$ \left\{ z \in \mathbb{C} \,\colon \left|z e^{1-z}\right| \leq 1 \,\,\text{ and }\,\, |z| \leq 1 \right\}. $$

As the proof is short and sweet I will include it here.

Proof: Let $z \in \mathbb{C}$ with $|z| \leq 1$ and $\left|z e^{1-z}\right| \leq 1$. Then

$$\begin{align*} \left|1 - e^{-nz} s_n(nz)\right| &= \left|e^{-nz} \sum_{k=n+1}^{\infty} \frac{n^k z^k}{k!} \right| \\ &= \left|\left(z e^{1-z}\right)^n e^{-n} \sum_{k=n+1}^{\infty} \frac{n^k z^{k-n}}{k!} \right| \\ &\leq e^{-n} \sum_{k=n+1}^{\infty} \frac{n^k}{k!} \\ &= 1 - e^{-n} s_n(n) \\ &&lt 1. \end{align*}$$

Having $s_n(nz) = 0$ here would contradict this inequality.

Q.E.D.

This bound is sharp in the sense that the limit points of the zeros of $s_n(n z)$ are precisely the points in the set

$$ \left\{ z \in \mathbb{C} \,\colon \left|z e^{1-z}\right| = 1 \,\,\text{ and }\,\, |z| \leq 1 \right\}. $$

This is known as the Szegő curve.

References:

[1] K. S. K. Iyengar, A note on the zeros of $\sum_{r=0}^{n} \frac{x^r}{r!} = 0$, The Mathematics Student 6 (1938), pp. 77-78.

(As the above paper apparently doesn't exist electronically I have transcribed it here.)

[2] J. D. Buckholtz, A characterization of the exponential series, The American Mathematical Monthly 73 (1966), no. 4, part II, pp. 121–123.

Here's another approach using Rouche's Theorem:

Let $f(z) = \frac{p_n(z)}{e^Z}$, and $g(z) = 1$. Note that $z \mapsto e^z$ has no zeros. Since convergence is uniform on compact sets (in this case $\bar B_r(0)$), we can choose $n$ large enough so that $|f(z) -1| &lt1$, $\forall z \in \bar B_r(0)$ (clearly $f$ has no zeros or poles on the circle of radius $r$). Then by Rouche's Theorem, $Z_f-P_f = Z_g-P_g$ (zeros and poles of $f,g$). Since $P_f=Z_g=P_g=0$, we have $Z_f = 0$, from which the result follows.