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,
Best Answer
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 $-b$?