TITLE: Zeros of MacLaurin polynomials for the exponential function
QUESTION [19 upvotes]: Asked but never answered at MSE.
Let $\exp_n(z)$ denote the nth degree Taylor polynomial of $e^z$ :
$\exp_n(z) = 1 + z + z^2/2! + ... + z^n/n!    \;$ .
The zeros of $\exp_n(z)$ were studied by Szego in the 1920's and 
later by others.   One of the consequences of Szego's results is
that the roots (after division by n) can come arbitrarily 
close to the imaginary axis.
Question: Is it possible for exp$_n$(z) to have a root that lies
precisely on the imaginary axis?

REPLY [16 votes]: Indeed the polynomial $\exp_n(z)$ has no purely imaginary zeros.  Write 
$$ 
\exp_n(ix) = C_n(x) + i S_n(x) 
$$ 
in the obvious notation, with $C_n$ for truncations of cosine, and $S_n$ for truncations of sine.   If there is a zero $ix$ for $\exp_n(z)$, then this $x$ must be a common root of $C_n$ and $S_n$, which are both polynomials in ${\Bbb Q}[x]$.  Therefore $C_n$ and $S_n$ must have some non-trivial gcd in ${\Bbb Q}[x]$.  Now we use the beautiful fact (going back to Schur) that $C_n(x)$ is irreducible, which completes our proof.  I learnt of this result of Schur from these lovely notes of Keith Conrad; simply apply Theorem 1 there, or see Corollary 1 which mentions this explicitly.