Nontrivial entire $f(z)$ never equal to $0$ [closed]

I'm looking for nonconstant entire functions $f(z)$ such that $f(z)\neq 0$ for any $z$.

More specifically I'm looking for nontrivial cases.

So $\exp(z),\exp(z^2),...$ is not what I am looking for.

I am aware that every solution is essentially equal to $\exp(g(z))$ for some entire $g$, but that does not mean every answer needs to be in that form.

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For instance f is never zero and f is given by An integral transform; so COMPOSITION of Some g and exp is not explicitly given. Or g has no closed form.

Hope this clarifies the confusion.

( in case a definition for entire f does not hold/converge everywhere you may use analytic continuation )

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Additionally I wonder if such a function can satisfy a simple functional equation of any kind.

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example

Assume f(z)= 2^z + 3^z - 7^z is never equal to 0.

Then ln(2^z + 3^z - 7^z) is entire and equal to g(z). F(z) = exp(g(z)).

Hope this clarifies.

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Edit 3

Maybe this related question makes it clear.

Why does $\sum a_i \exp(b_i)$ always have root?

This question is a generalisation of it. Hope it finally helps before this gets closed.

closed as unclear what you're asking by Jack D'Aurizio, Jonas Meyer, user147263, user99914, hardmathJul 4 '15 at 1:33

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• How in the world is $\exp(z)$ a "trivial" case? – 5xum Jul 3 '15 at 11:29
• The entire functions that never attain the value $0$ are exactly the functions of the form $z\mapsto\exp (g(z))$ with an entire $g$. – Daniel Fischer Jul 3 '15 at 11:30
• Guys plz i know that. I Said so myself in the op. – mick Jul 3 '15 at 12:17
• @mick: so why do you keep asking ? – Yves Daoust Jul 3 '15 at 12:32
• Hope my edit clarified the confusion. Functions can be given in many equivalent ways !! – mick Jul 3 '15 at 13:47

I am aware that every solution is essentially equal to $\exp(g(z))$ for some entire $g$, but that does not mean every answer needs to be in that form.

Yes, every answer needs to be in that form. The only theorem you need is that an entire function has a Taylor series with infinite radius of convergence, so integrating it term by term gives again a Taylor series with infinite radius of convergence, which so is an antiderivative of the given entire function. You also need the easy fact that for every $w\in\mathbb{C}$, $w\ne0$, there exists $w_0\in\mathbb{C}$ such that $\exp(w_0)=w$.

Let $f$ be an entire function that has no zeros. Then $$h(z)=\frac{f'(z)}{f(z)}$$ defines an entire function. Let $g(z)$ be an antiderivative of $h(z)$ and consider $$F(z)=f(z)\exp(-g(z))$$ which is entire as well. Then \begin{align} F'(z)&=f'(z)\exp(-g(z))+f(z)(-g'(z)\exp(-g(z)))\\[6px] &=f'(z)\exp(-g(z))-f(z)\frac{f'(z)}{f(z)}\exp(-g(z))\\[6px] &=0 \end{align} Thus $F$ is constant and $F(z)=F(0)=f(0)\exp(-g(0))$, so $$f(z)=f(0)\exp(-g(0))\exp(g(z))$$ Since we can choose whatever value we want for $g(0)$ and $f(0)\ne0$, we can assume that $$\exp(g(0))=f(0)$$ and, for such $g$, we have $$f(z)=\exp(g(z))$$ for all $z\in\mathbb{C}$.

So, indeed, any entire function that does not assume the value $0$ is of the form $\exp(g(z))$ for some entire function $g$.

By the Weierstrass factorization theorem, an entire function that has no zeros must be of the form $\exp(g(z))$.

• I Said that in the op already. – mick Jul 3 '15 at 12:17
• @mick if you already know this what was the point in asking ?? Both answers are telling you that every function satisfying your condition must be on that form, you seem to disagree – Alonso Delfín Jul 3 '15 at 13:05
• No i do not disagree. But a function can be of the form exp(g(z)) but given as say An integral transform such that exp composition is not given implicit, and g has no closed form. See ? – mick Jul 3 '15 at 13:37
• I made 2 edits to clarify, hope it helped. – mick Jul 3 '15 at 18:35
• I made another edit to clarify. Hope it helps. Getting desperate :( Dont close. – mick Jul 6 '15 at 16:48