# Where are the zeroes of complex exponential function?

There is none because we have $e^z \neq 0$ for every $z \in \mathbb C$.

But we have Taylor series that everywhere converges to $e^z$, it is $e^z = \displaystyle \sum_{k=0}^{+ \infty} \frac {z^k}{k!}$.

If we truncate that series , say, at natural $m$, then we have Taylor polynomial $\displaystyle \sum_{k=0}^{m} \frac {z^k}{k!}$, which has, counted with maybe possible multiplicity, $m$ complex zeroes.

So as the degree of Taylor polynomial grows the number of zeroes increases, but in the limit they all dissapear, why?

• Why would you expect otherwise? – anomaly Aug 27 '18 at 3:47
• Since the Taylor polynomial $T_n(z)$ tends to $e^z$ uniformly for $|z| < R$ for any fixed radius $R$, and $e^z$ has no zeroes, for large enough $n$ the zeroes of $T_n(z)$ must have $|z| > R$. In other words - the zeroes must tend to infinity. It would be interesting to see what the zeroes look like on the complex plane for large $n$. – Jair Taylor Aug 27 '18 at 3:56
• @JairTaylor Do you think that they form some special polygons or something like that? – Right Aug 27 '18 at 4:01
• @anomaly Because there are reasons why the number of zeroes is not infinite. – Right Aug 27 '18 at 4:02
• @Right Yes, the zeroes of polynomial sequences often form interesting shapes. – Jair Taylor Aug 27 '18 at 4:06

Nice question. A google search brings me to a paper of Ian Zemke, although this does not seem to have been peer-reviewed and I have not checked it myself. He apparently proves that the zeroes of $T_n(z) = \sum_{k=0}^n\frac{z^k}{k!}$ tend toward a specific curve, after normalization. That is, if $p_n(z) = T_n(nz)$, then the zeroes of $p_n(z)$ are asymptotically close to the curve $$Γ = \{z : |ze^{1−z} | = 1, |z| ≤ 1\}.$$ Thus the zeroes of the original $T_n(z)$ tend to infinity linearly with $n$.