Solution 11. By Hadamard's theorem if it omits $a$ and $b$, then we have $f(z)-a=e^{p(z)}$ and $f(z)-b=e^{q(z)}$ for polynomials $p$ and $q$. Then $e^{p(z)}-e^{q(z)}=C$ for some constant $C$. Letting $z$ tend to infinity we see that the leading terms of the polynomials $p$ and $q$ must be the same. Say it is $a_n z^n$. Then, considering the limit as $z$ tends to infinity of \[ e^{p(z)-a_n z^n}-e^{q(z)-a_n z^n}=C e^{-a_n z^n}, \] we see that the next leading terms are also equal. Proceeding by induction shows that $p=q$, but then this is a contradiction since it would imply that $b=a$. But we assume they are distinct.