# Complex Analysis: #29 The Order of an Entire Function

**Theorem 49**Let f : ℂ → ℂ be an entire function. Assume that there exist real constants C > 0, λ > 0, such that Re(f(z)) ≤ C(1 + |z|

^{λ}), for all z ∈ ℂ. Then f is a polynomial, at most of degree [λ].

*Proof*

To begin with, we know that, for k,

*l*∈ Z, we have

Therefore, taking r → ∞, we see that if k > λ then a

_{k}= 0. An analogous argument shows also that b

_{k}= 0.

**Deﬁnition 17**

An entire function f is said to have

*ﬁnite order*if there exists some real number ρ > 0 and a constant C > 0, such that

|f(z)| ≤ Ce

^{|z|}^{ρ},for all z ∈ ℂ. The inﬁmum over all such ρ is the order of f. That is to say, α is the order of f if |f(z)| ≤ Ce

^{|z|}

^{}

^{α+∈}for all ∈ > 0 and z ∈ ℂ. If |f(z)| ≤ Ce

^{|z|}

^{α}for all z ∈ ℂ then α is the

*strict order*of f.

**Theorem 50**Let f be an entire function of ﬁnite order with no zeros. Then f = e

^{g}, where g is a polynomial whose degree is the order of f.

*Proof*

According to exercise 12.1, given f, then there exists an entire function g with f = e

^{g}. But then we must have

So Re(g(z)) ≤ |z|

^{α}if α is the order of f, and therefore the result follows from theorem 49.

Going beyond this, we would like to think about entire functions of ﬁnite order, but

*with*zeros This leads us to Hadamard’s theorem. But before we arrive there, let us think about Jensen’s formula.

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