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I am struggling with one exercise here. It says:

Let $f(z)$ be an entire function, and the set of zeros of $f$ is finite, but nonempty. Show that $f$ takes all complex values.

My idea is: Assuming there is some $b$ s.th. there is no $a$ with $f(a)=b$, then define the function $g:=f-b$, so then $g$ will be entire (since it's an entire function just shifted), and $g$ will never be zero. But that is a contradiction since an entire function can be written by definition as a polynomial, and any polynomial has a root somewhere, so $g=0$ somewhere, so $f=b$ somewhere, contradiction.

But there is little Picard's theorem that tells us an entire function omits at most one value. Now suddenly I seem to show that it omits no value. And the fact that the set of zeros is finite but nonempty is only used, basically, to say that $f$ is not constant in the first place. So where is the mistake?

Best regards,

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4  
"An entire function can be written by definition as a polynomial" -- This step is, of course, false. The exponential function is entire but not a polynomial. –  Srivatsan Dec 15 '11 at 11:44
    
en.wikipedia.org/wiki/Entire_function For who, like me, is far from the last course in Compolex Analysis. –  Giovanni De Gaetano Dec 15 '11 at 12:10
    
Hmm, what I meant is that we can write an expansion for the entire function from which the zeros can be seen? But either way, isn't it true that an entire function has a zero somewhere? –  Marie. P. Dec 15 '11 at 12:12
    
The function $e^z$ is entire and has no zero. There is a theorem that says that if $f$ is entire and has no zero then $f=e^g$ for some entire function $g$. –  Gerry Myerson Dec 15 '11 at 12:16
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Interesting -- this means that Picard's little theorem can be slightly extended to say that a non-constant entire function takes all values or takes all but one value infinitely often. –  joriki Dec 15 '11 at 12:42

1 Answer 1

up vote 3 down vote accepted

Hint: Suppose $f(z)$ never takes the value $b$ and write $f(z)-b=e^{g(z)}$ for some entire function $g$. The left hand side of the equations takes the value $-b$ only finitely many times. How many times does the right hand side take the value $-b$?

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