Each polynomial of degree $n$ has $n$ different roots.
We know that $$e^z = \sum_{n=0}^\infty \frac {z^n}{n!}$$ has no roots. What is the behaviour then of the root of $S_N = \sum_{n=0}^N\frac {z^n}{n!}$? Where do they go?
We could generalize this and ask
Given $f(z)$ holomorphic in some $\Omega$ (feel free to take any condition you want on $\Omega$), what is the relationship between the root of $f(z)$ and the roots of the truncated expansion of $f(z)$?
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I've seen the suggested duplicate and its really interesting, though I am interested in a more general result :-)
The roots go off to infinity. One can show the roots of $\displaystyle f(z) = \sum_{k=0}^{N-1} \frac{z^k}{k!}$ tend to a curve $|z e^{1-z}| = 1$ if you rescale by a factor of $N$.
Please see
I even wrote a paper on the topic myself on the roots of truncated cosine. [1] Here's a chart I produced for this occasion where you can see the linear outward drift of the roots.
Here is the numpy code I used to produce it. Mathematica or Pari might be better suited.