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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 $b$ 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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"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 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
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
up vote 4 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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-@Johan in your proof is necessary to use Picard's little theorem or entire functions have dense images? – felipeuni Aug 27 '14 at 19:14
I think Picard's little theorem basically is needed. A slight weakening, which states that an entire function omits at most a finite number of values, should be enough though. If $g$ just took values on a dense set, it could still avoid the values where $e^g(z)=-b$. But knowing that $g(z)$ can omit at most at a finite number of values it is clear that it must take of values of the form $\ln(-b)+ k 2\pi i$ infinitely many times, where the branch of the logarithm is arbitrary. – Johan Sep 15 '14 at 13:00

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