Well, it looks like Office Shredder is as foolish as possible -- because for
every nonzero polynomial f(z), the function e^z - f(z) has infinitely many zeros in the complex plane.
Proof: Suppose that e^z - f(z) has only finitely many zeros, say, z_1, z_2, \ldots, z_n (where the zeros are listed with their multiplicity). Let g(z) = \prod_{j=1}^{n} (z-z_j). Then after removing the removable singularities, we have that:
\frac{e^z - f(z)}{g(z)}
is an entire function which is never zero, and hence is equal to e^{h(z)} for some entire function h. Now, observe that for all sufficiently large z, |g(z)| \geq 1 and |f(z)| \leq e^{|z|}. Therefore for all sufficiently large z:
e^{\Re (h(z))} = |e^{h(z)}| = \left| \frac{e^z - f(z)}{g(z)} \right| \leq |e^z - f(z)| \leq |e^z| + |f(z)| \leq 2e^{|z|} \leq e^{2|z|}
And hence for all sufficiently large z:
\Re(h(z)) \leq 2|z|
By applying the
Borel-Caratheordory theorem, it follows that |h(z)| grows at most linearly with |z|, which in turn implies that h(z) must actually be a linear polynomial. That is, for some complex constants a and b, we have h(z) = az + b.
Now, recall that we defined h(z) such that:
\frac{e^z - f(z)}{g(z)} = e^{h(z)} = e^{az+b}
So:
e^z = f(z) + e^{az+b}g(z)
And thus:
1 = f(z)e^{-z} + e^{(a-1)z + b}g(z)
Take limits of both sides as z \rightarrow \infty along the positive real axis. Clearly f(z) e^{-z} \rightarrow 0. Since the left hand side is a constant 1, this implies that e^{(a-1)z + b}g(z) \rightarrow 1. Now, if \Re(a-1) < 0, then as z \rightarrow \infty along the positive real axis, we have that e^{(a-1)z + b}g(z) \rightarrow 0, and if \Re(a-1) > 0, then e^{(a-1)z + b}g(z) \rightarrow \infty. Therefore \Re(a-1) = 0, and so |e^{(a-1)z + b}| is constant on the real line. Since |e^{(a-1)z + b}g(z)| \rightarrow |1| = 1 as z \rightarrow \infty along the positive real axis, this implies that |g(z)| approaches a nonzero constant as z \rightarrow \infty along the positive real axis. This is only possible if g has degree zero -- i.e., g is a constant. Letting c = e^b g, we can rewrite our equation as:
1 = f(z)e^{-z} + ce^{(a-1)z}
We still have that the second term on the RHS approaches 1 as z \rightarrow \infty along the positive real axis. We also know that \Re(a-1) = 0 -- i.e. a-1 is pure imaginary. Now, if a-1 is a nonzero imaginary number, e^{(a-1)z} will oscillate and fail to approach a limit as z \rightarrow \infty along the positive real axis. Therefore, a-1 = 0. Which means the second term on the RHS is just the constant c, and so we must have c=1. Hence:
1 = f(z)e^{-z} + 1
Which by simple algebra, implies that f(z) = 0. Q.E.D.