About canonical factors for Weierstass infinite products.

Question

I was reading the proof of below theorem (p.145 complex analysis Elias M.Stein):
Given any sequence $\{a_n\}$ of complex numbers whit $|a_n| \to \infty $ as $n \to \infty $, there exists an entire function $f$ that vanishes at all $z=a_n$ and nowhere else.
In the proof writer introduces canonical factors of degree k as: $E_k(z) = (1-z)e^{z+\frac{z^2}{2}+ \ldots +\frac{z^k}{k}}$ .
My question: What is canonical about these factors?

Answer 1

In the proof of the theorem, we use products of $E_k(z\cdot z_k^{-1})$, where the $z_k$ are the prescribed zeros. So we want a function $E_k$ which is $0$ at $1$, which explains why the factor $1-z$ appears. Indeed, if we can find a function which vanishes at $1$, it will be possible for any point.

But we also want convergence of the product $\prod_kE_k(z\cdot z_k^{-1})$ (with different indexes if we want other multiplicities), so we have to find a function $f_k(z)$ in order to make $E_k(z)=(1-z)f_k(z)$ near $1$ as $k\to +\infty$. When $z$ is real, we have that $f_k$ is the exponential of the partial sum of order $k$ of the series of $-\log(1-z)$.

Actually, in order to ensure the convergence, we have use the result $|1-E_k(z)|\leqslant |z|^{k+1}$ when $|z|\leqslant 1$.