Find minimum $n$ such that $1+z+\frac{z^2}{2!}+\cdots+\frac{z^n}{n!}=0$ has no answer inside the circle of radius $100$ centered at the origin What is the minimum $n$ for $$1+z+\frac{z^2}{2!}+\cdots+\frac{z^n}{n!}=0$$ such that there is no solutions in the disk of radius $100$ around the origin?
please give me a hint.
 A: This is relevant: complex zeros of the polynomials $\sum_{k=0}^{n} z^k/k!$ inside balls
As pointed out in the comments, Rouché's theorem is your best bet for this kind of problems, since it tells us when two functions have the same number of zeros (counting multiplicities) in an open set. It goes like this:

Let $U\subseteq\mathbb{C}$ be an open set and $f$,$g$ holomorphic functions in $U$. Let $V\subset\subset U$ such that $\partial V =\Gamma$ is a cycle (I couldn't find the technical term for this!). Then, if $|f-g|<|f|$ in $\Gamma$, $f$ and $g$ have the same amount of zeros in $V$ counting multiplicities.

This is just one of the many forms this theorem has. For your case, let's consider $U=\mathbb{C}$, $V=D_{100}(0)$ and
$$
f(z) = e^{z}
$$
$$
g(z) = 1+z+\frac{z^2}{2!}+\cdots+\frac{z^n}{n!}.
$$
So, for any $z$ such that $|z|=100$,
$$
|f-g| = \left\lvert \sum_{k=n+1}^\infty \frac{z^k}{k!}\right\rvert \leq \sum_{k=n+1}^\infty \frac{100^k}{k!} 
$$
This sum is exactly the remainder of the Taylor series of the real exponential, which is given by:
$$
R_{n}(100) = \frac{1}{n!}\int_0^{100} (100-t)^{n}e^t dt = \frac{e^{100}}{n!}\int_0^{100} x^n e^{-x}dx. 
$$
Finally, observe that $e^{-100}\leq |f|$ in the domain we are considering. The $n$ you need is precisely the one for which $R_n(100)< e^{-100}$, or,
$$
\frac{1}{n!}\int_0^{100} x^n e^{-x}dx < e^{-200} 
$$
From this point on, I can only think about estimating the result, maybe someone else could help us solve this analytically.
