# Finding alternate form of entire function?

So we were given a problem and it follows some of the stuff we did with Picard's theorems in class, but we were never shown how to actually find the form, just what they looked like for a few cases like not equal to a constant, or not equal to z.

f(z) is entire and $$f(z)\ne\frac{1}{z}$$ for $$z\ne0.$$ We have to show that this can be rewritten as $$\frac{1-e^{zh(z)}}{z}$$ where h(z) is entire. I have no clue where to get started. I know we would use the fact the function is entire and you have a branch of the Log, but not sure how to use that in this case.

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## 1 Answer

$F(z)\ne1/z$ is the same as $1-zf(z)\ne0$; $f(z)$ entire implies $1-zf(z)$ entire; so we have an entire function $1-zf(z)$ which is never zero. I take it you know this implies $1-zf(z)=e^{g(z)}$ for some entire $g$, so you get $$f(z)={1-e^{g(z)}\over z}$$ Now $g(0)$ had better be $2\pi in$ for some integer $n$, lest this formula force an unwanted pole at zero on $f$, so $g(z)=2\pi in+zh(z)$ for some entire $h$, and away we go.

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Ohhhh. Man did I miss that. I was trying to re-write it as zf(z)=1 and go from there rather than set it equal to 0. Guess that's what happens when you stare at a problem forever. Thanks a lot! Makes prefect sense now. –  Roy Apr 22 '12 at 11:14
If your question is now answered, you can "accept" the answer by clicking in the little check mark next to the answer. –  Gerry Myerson Apr 22 '12 at 13:09