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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?

Right
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    Why would you expect otherwise? – anomaly Aug 27 '18 at 03:47
  • 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 03:56
  • @JairTaylor Do you think that they form some special polygons or something like that? – Right Aug 27 '18 at 04:01
  • @anomaly Because there are reasons why the number of zeroes is not infinite. – Right Aug 27 '18 at 04:02
  • @Right Yes, the zeroes of polynomial sequences often form interesting shapes. – Jair Taylor Aug 27 '18 at 04:06
  • At 11 PM, we put 10 balls numbered 1--10 into a bag, then remove the ball numbered 1. At 11:30 PM, we put balls numbered 11--20 into a bag, then remove number 2. At 11:45 PM, put 21--30 into the bag, and remove 3. Repeat this procedure (at half the time to midnight, add ten balls, then remove the ball with the lowest number). At midnight, how many balls are in the bag? It turns out that the "right" answer is zero, since every ball is eventually removed before midnight. At any finite step, there are more balls in the bag than in any previous step, but they are all gone by midnight. – Xander Henderson Aug 27 '18 at 04:08
  • You might think of something similar going on here. Any finite approximation has a lot of zeros, but those zeroes are farther and farther out as the approximations get better. The zeroes are being "pushed out" to infinity. – Xander Henderson Aug 27 '18 at 04:08
  • Related. See also the questions under "Linked" on the sidebar there. – Antonio Vargas Sep 03 '18 at 21:01

1 Answers1

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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 $n$.

A nice visualization from Zemke's paper:

enter image description here

Edit: It seems this result is actually due to Gábor Szegő from 1924. See Hans Lundmark's comment below.

Jair Taylor
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