Bounds for Weierstrass Elementary Factors

Theorem
Let $E_p: \C \to \C$ denote the $p$th Weierstrass elementary factor:


 * $E_p \left({z}\right) = \begin{cases} 1 - z & : p = 0 \\

\left({1 - z}\right) \exp \left({z + \dfrac {z^2} 2 + \cdots + \dfrac{z^p} p}\right) & : \text{otherwise}\end{cases}$

Let $z \in \C$.

Some bound
Let $\left\lvert{z}\right\rvert \le \dfrac 1 2$.

Then:
 * $\left\lvert{E_p \left({z}\right) - 1}\right\rvert \le 3 \left\lvert{z}\right\rvert^{p + 1}$

Another bound
Let $\left\lvert{z}\right\rvert \le 1$.

Then:
 * $\left\lvert{E_p \left({z}\right) - 1}\right\rvert \le \left\lvert{z}\right\rvert^{p + 1}$

Proof of some bound
Let $\left\lvert{z}\right\rvert \le \dfrac 1 2$.

We may assume $p \ge 1$.

We have:
 * $E_p \left({z}\right) = \exp \left({\log \left({1 - z}\right) + \displaystyle \sum_{k \mathop = 1}^p \frac{z^k} k}\right)$

Then:

Because $p \ge 1$:
 * $2 \left\lvert{z}\right\rvert^{p + 1} \le \dfrac 1 2$

By Bounds for Complex Exponential:
 * $\left\lvert{E_p \left({z}\right) - 1}\right \rvert \le 3 \left\lvert{z}\right\rvert^{p + 1}$

Also see

 * Weierstrass Factorization Theorem, what this is made for
 * Bounds for Complex Exponential
 * Bounds for Complex Logarithm