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Page 69 of $\textbf{[1]}$, one finds the theorem 3.8 (Tumura-Hayma):

Suppose that $~f(z)~$ is meromorphic and has only a finite number of poles in the plane, and that $~f(z)~$ and $~f^{(l)}(z)~$ have only a finite number of zeros for some integer $~l\geq 2.~$ Then : $$f(z)=\dfrac{P_1(z)}{P_2(z)}e^{P_3(z)}$$ where $~P_1,~P_2,~P_3~$ are polynomials. If, further, $~f(z)~$ and $~f^{(l)}(z)~$ have no zeros, then $$f(z)=e^{Az+B}~~~~or~~~~f(z)=\dfrac{1}{(Az+B)^n}$$ where $~A,~B~$ are constants such that $~A\neq 0~$ and $~n~$ is a positive integer.}

$\textbf{QUESTION}$ : May we hope for a ``simple" proof of the following particular case of 3.8 (which is also theorem 5, page 22 of $\textbf{[2]})$ ?

Suppose that $~f(z)~$ is an entire function, and that $~f(z)~$ and $~f^{(2)}(z)~$ have no zeros in the plane. Then $~f(z)=e^{Az+B}~$ where $~A,~B~$ are constants such that $~A\neq 0.~$

$\textbf{[1]}$ ``Meromorphic Fuctions", W.K. Hayman, 1964, Oxford Mathematical Monographs

$\textbf{[2]}$ ``Picard values of Meromorphic Functions and their Derivatives", W.K. Hayman, Annals of Mathematics, Second Series, Vol. 70, No 1 (Jul.,1959), pp. 9-42

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Let $p$ be the order of $f$, and let $k = \lfloor f \rfloor$. Since $f$ is entire and never vanishes, Hadamard’s theorem implies $f = e^{g(z)}$ where $g$ is a polynomial of degree $k$. Hence, $f'(z) = g'(z)e^{g(z)}$. Now, the fact that $f'$ never vanishes implies that $g'(z)$ must be constant, since a nonconstant polynomial must vanish at some point. Therefore, $g(z) = az + b$ for some constants $a, b \in \mathbb{C}$, so $f = e^{az + b}$. – Sanath K. Devalapurkar Mar 28 '14 at 1:04
What if $p=\infty$? There was no assumption that $f$ is of finite order. – Noam D. Elkies Mar 28 '14 at 1:09
@NoamD.Elkies That is indeed true. If $p=\infty$, then the above argument does not work, since $g(z)$ would also be a polynomial of degree $\infty$. I should have said that if and only if $f$ was of finite order the argument would work. P.S. Clarification in my first comment - I meant $k=\lfloor p\rfloor$, not $k=\lfloor f\rfloor$. – Sanath K. Devalapurkar Mar 28 '14 at 1:22

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