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