Define $f_n:\mathbb{C}\to\mathbb{C}$ and $(\alpha_n)$ such that:$$f_n(z)=\sum_{k=0}^n \frac{z^k}{k!}$$ and $f_n(\alpha_n)=0$. Prove $|\alpha_n|\to\infty$ as $n\to\infty$.
I guess this makes sense since $\exp z$ has no roots, but I'm a little stuck on this. I thought of two ideas:
to consider $$N=\frac{1}{2\pi i}\int_{\gamma}\frac{f_n'(z)}{f_n(z)}\,\mathrm{d}z=\frac{1}{2\pi i n! }\int_{\gamma} \frac{z^n}{f_n(z)}\,\mathrm{d}z$$ where $N$ is the number of zeros of $f_n$ inside $\gamma$, an arbitrary circle centered at the origin. My strategy would then be to show that for $n$ large, $N=0$.
to somehow use Rouche's theorem, so that we can deal with a simpler function for which the result might be more obvious.
Are any of these worth pursuing? Any help would be appreciated.
It's much easier than you think. $\exp$ maps straight lines parallel to the imaginary axis to circles centered at the origin. The radius of the circle is exactly $\exp$ of the number that is the intersection of the line with the real axis. In particular, it implies that for $|z| < R$, $|\exp(z)| > e^{-R}$.
Now, take any $R > 0$. We'll show that for large enough $n$, $|\alpha_n| > R$, which implies that $\alpha_n \to \infty$.
For $|z| < R$ We have:
$$ |\exp(z)-f_n(z)| = \left|\sum_{k=n+1}^\infty \frac{z^k}{k!} \right| \leq \sum_{k=n+1}^\infty \frac{R^k}{k!} $$
Since $\sum_{k=0}^\infty \frac{R^k}{k!}$ is convergent, its tail $\sum_{k=n+1}^\infty \frac{R^k}{k!}$ is arbitrarily small for large enough $n$. Pick $n$ such that
$$ \sum_{k=n+1}^\infty \frac{R^k}{k!} \leq \frac{e^{-R}}{2} $$
Since $|\exp(z)| > e^{-R}$ for $|z| < R$, and $|\exp(z)-f_n(z)| \leq \frac{e^{-R}}{2}$, we necessarily have $|f_n(z)| > \frac{e^{-R}}{2}$, so $f_n(z)$ has no zeroes with $|z| < R$ for large enough $n$.