Order of Shifted Entire Function

Theorem
Let $f: \C \to \C$ be an entire function of order $\alpha$.

Let $a \in \C$.

Then $f \left({z + a}\right)$ has order $\alpha$.

Proof
We shall verify.

Let $\map g z := \map f {z + a}$ for $z \in \C$.

Then for all $R > \cmod a$:
 * $\ds \max_{\cmod z \le R - \cmod a} \cmod {\map f z} \le \max_{\cmod z \le R} \cmod {\map g z} \le \max_{\cmod z \le R + \cmod a} \cmod {\map f z}$

Thus:
 * $\paren 1 : \dfrac {\ds \ln \ln \max_{\cmod z \le R - \cmod a} \cmod {\map f z} }{\ln R} \le \dfrac {\ds \ln \ln \max_{\cmod z \le R} \cmod {\map g z} }{\ln R} \le \dfrac {\ds \ln \ln \max_{\cmod z \le R + \cmod a} \cmod {\map f z} }{\ln R}$

On the other hand, for each $c \in \R$:
 * $\ds \paren 2 : \lim_{R \to +\infty} \dfrac {\map \ln {R + c} } {\ln R} = 1$

Thus:

Similarly, we have:
 * $\ds \paren 4 : \lim_{R \to +\infty} \dfrac {\ds \ln \ln \max_{\cmod z \le R + \cmod a} \cmod {\map f z} }{\ln R} = \alpha$

In view of $\paren 3$ and $\paren 4$, the claim follows from $\paren 1$ by Sandwich Rule.

Also see

 * Order of Product of Entire Functions