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Prove that, for each $R > 0$, there exists $N(R) \in \mathbb N$, such that, for all $n \geq N(R)$, the function $$f_n(z) = \sum_{k=0}^{k=n} \frac {z^k}{k!}$$ has exactly $n$ zeros in $|z| < R$.

This question is asked after Rouche's theorem, so I assume I have to apply the theorem to prove this.

Rouche's theorem : If $f$ and $h$ are each functions that are analytic inside and on a simple closed contour $C$ and if the strict inequality $|h(z)|<|f(z)|$ holds at each point on $C$, then $f$ and $f+h$ must have the same total number of zeros inside C.

I would like to get some hint on how to prove this.

ab123
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  • See https://math.stackexchange.com/questions/535720/what-are-the-properties-of-the-roots-of-the-incomplete-finite-exponential-series and links contained in the answers. – Lutz Lehmann Nov 13 '17 at 20:34
  • @Lutz the results there seem to be more advanced than what is required to prove here, is there a simpler way to do something here? – ab123 Nov 13 '17 at 20:40
  • Are you sure of this question? The roots of $f_n(nw)$ accumulate at a bounded teardrop shape, thus the roots of $f_n(z)$ have a magnitude proportional to $n$. The question should be that all roots are outside this disk, or in $|z|>R$. And this question has been asked and answered before. – Lutz Lehmann Nov 13 '17 at 20:50

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