Why does $\sum a_i \exp(b_i)$ always have root? Let $z$ be complex.
Let $a_i,b_i$ be polynomials of $z$ with real coefficients.
Also the $a_i$ are non-zero and the non-constant parts of the polynomials $b_i$ are distinct.
Let $j > 1$.
$$f(z) = \sum_{i=1}^j a_i \exp(b_i)$$
It seems there always exists a complex value $s$ such that
$$ f(s) = 0$$
Is this true?
If so, why? 
How to prove this? If false, what are the simplest counter-examples?
 A: This follows from the theory of entire functions of finite order
in complex analysis.  Specifically, we have:
Proposition:
Suppose $f(z) = \sum_{i=1}^j a_i \exp b_i$ for some polynomials $a_i,b_i$
(which may have complex coefficients, though the question specifies
real polynomials).  Then if $f\,$ has no complex zeros then there exists
a polynomial $P$ such that $f = \exp P$. 
Proof: let $d = \max_i\max(\deg a_i,\deg b_i)$.  If $d \leq 0$ then $f$ is constant
and we may choose for $P$ a constant polynomial.  Else there exists
a constant $A$ such that $\left|\,f(z)\right| \leq \exp(A\left|z\right|^d)$
for all complex $z$.  This makes $f$ an
entire
function of order at most $d$.
If $f$ has no zeros then $f = e^g$ for some analytic function $g$,
and it follows that $g$ is a polynomial (by a special case of the 
Hadamard product for an entire function of finite order). $\Box$
Moreover, once we put the expansion $f(z) = \sum_{i=1}^j a_i \exp b_i$
in normal form by assuming that each $b_i$ vanishes at zero
(else subtract $b_i(0)$ from $b_i$ and multiply $a_i$ by $e^{b_i(0)}$),
then at least one of the $b_i$ is $P-P(0)$, and we can cancel and
combine terms to identify $f$ with $\exp P$.  The proof (by considering
behavior for large $|z|$) is somewhat tedious, though much easier in
the real case [hint: start by writing $f(z) \, / \exp P(z)$ as 
$\sum_{i=1}^j a_i \exp (b_i-P)$].  In particular, if $j>1$ and
no two $b_i$ differ by a constant then $f$ cannot equal $\exp P$
and thus must have complex zeros.
