There is a well known fact that if one considers the family of polynomials $$f\_n(z) = \sum\_{k=1}^n \frac{z^k}{k!}$$ that any sequence of solutions to $f\_n(nz\_n)=0$ must converge to a point $z$ on the curve $|ze^{1-z}|=1.$ For reference, this curve is called the Szego curve. 


My question is about the convergence of the roots. Are the roots proved to be exactly on the curve or near the curve and approaching it?



Do the roots lie exactly on the Szego Curve or Approach it?
I assume you meant "if $z\_n$ satisfies $f(n z\_n)=0$ and if the sequence $z\_n$ converge to a nonzero limit $z$ then $|z e^{1-z}|=1$ ".


Now, if $n z\_n$ is a root of $f\_n$ then $z\_n$ is algebraic over $\mathbb{Q}$. Assume $z\_n$ also lies on your curve. Its real part is a sum of two algebraic numbers, hence is algebraic, and so is $\alpha := 1 - Re(z\_n)$. If $\alpha = 0$ then $|z\_n| = Re(z\_n) = 1$ so that $z\_n = 1$, but this would contradict $f(n z\_n) = 0$. Thus $\alpha \neq 0$, and the algebraic number $z\_n \times \bar z\_n = e^{-2 \alpha}$ must be transcendental, a contradiction. 


