 Solution

Again suppose $P$ is a nonconstant polynomial without a complex zero, and put $g=1/P$. As proved in part (c), $g$ is an entire holomorphic function with $|g(z)|\to 0$ at infinity. Choose $z_0$ and then a disk large enough that $|g(z)|<|g(z_0)|/2$ outside it. On the closed disk, $|g|$ attains its maximum by the extreme value theorem, and that maximum is positive and greater than every value outside the disk. It is therefore a global, and in particular local, maximum on $\mathbb{C}$.

Part (b)(ii) forces $g$ to be constant. A nonzero constant reciprocal makes $P$ constant, a contradiction. Hence $\boxed{P\text{ has a complex zero}}$. Equivalently one can minimize $|P|$ globally and apply the maximum modulus principle to its reciprocal if that minimum is positive. This supplies the second proof of the fundamental theorem of algebra using part (b).