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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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    "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.2011-12-15
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    http://en.wikipedia.org/wiki/Entire_function For who, like me, is far from the last course in Compolex Analysis.2011-12-15
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    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?2011-12-15
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    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 $g$.2011-12-15
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    @Marie Reg your last comment, I think you're referring to the [Weierstrass factorization theorem](http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem), but I am not sure. And, it is not true that an entire function must be zero somewhere, the exponential function being a counterexample. [More generally, it is possible for an entire function to miss any one point $a \in \mathbb C$. Consider the function $e^z+a$.]2011-12-15
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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.2011-12-15

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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](http://en.wikipedia.org/wiki/Liouville%27s_theorem_(complex_analysis)#Entire_functions_have_dense_images)?2014-08-27
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    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.2014-09-15