Is there any nice explanation of why the complex exponential function has no roots in the complex plane?

Question

Here I am not looking for an explanation that uses basic properties that complex exponential function has, such as $e^{z+w}=e^ze^w$ or $e^0=1$ or any other, if this fact can be explained by using those basic properties.

I am seeking for some explanation that has to do with the positions and number of the roots of the truncated exponential function and suppose that we only know how Taylor series for $e^z$ looks like, so in fact I seek for an explanation in which we do not know that complex exponential function has the basic properties it has.

Suppose that we truncate complex exponential function and define a function $e_{k}{(z)}=\sum_{i=0}^{k} \dfrac {z^i}{i!}$.

Because of the fundamental theorem of algebra we have that $e_{k}{(z)}$ has $k$ complex roots so bigger the $k$ the more roots we have.

But when we pass to the limit $\lim_{k\to\infty} e_{k}{(z)}=e^{z}$ somehow all the roots "disappear", and instead of maybe expected an infinite number of roots we have none.

How to explain this?

Answer 1

Quoting from Locating the zeros of partial sums of $\exp(z)$ with Riemann-Hilbert methods:

We denote by $p_{n}(z) := 1 + z + \cdots + \frac{z^{n}}{n!}$ the partial sums of the exponential series. The problem to describe the asymptotic distribution of the zeros of $p_n$ was posed and solved in the classical paper of Szegő [11]. He proved that the zeros of $p_n$, divided by $n$, converge in the limit $n \to \infty$ to some curve $D_{\infty}$, now called Szegő curve, which consists of all complex numbers $|z| \leq 1$ that satisfy the equation $|z e^{1-z}| = 1$.

So, it seems that the zeros of $p_n$ go to infinity as $n$ increases, leaving no zeros for $\exp$.

Here is a precise statement, paraphrasing this answer:

Every zero of the polynomial $\displaystyle s_n(z) = \sum_{k=0}^{n} \frac{z^k}{k!}$ lies in the annulus $\displaystyle \frac{n}{e^2} < |z| < n.$

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