6
$\begingroup$

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,

  • 1
    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

1 Answers 1

4

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 $-b$?

  • 0
    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