= 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 $|g(z)|\to0$ 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).
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