The site's wording fails for every $P$. Since $P(0)=1$, the point $0$ is not in $\{z:\lvert P(z)\rvert<1\}$, so no path inside that set starts at $0$, and the answer is no for every $C$, trivially; the smallest instance is $P(z)=1-z$. It fails a second way at the constant polynomial $P=1$, which has $P(0)=1$ and no roots: its set is empty, so no path exists even with the starting point exempt. The change inserts "everywhere except at $z=0$" after the set and "nonconstant" before "polynomial"; nothing else changes. The first insertion is in the posers' own words. Erdős, Herzog and Piranian [EHP55, §1, p. 347] (library card: Erdős, Herzog and Piranian 1955) state Cohen's theorem as giving a path from the origin to the unit circle on which "the inequality $|P| < 1$ holds everywhere except at $z = 0$", and two paragraphs later ask, in connection with their Theorem 1, whether a universal constant $L$ exists such that for every polynomial (1) "the inequality $|P| < 1$ holds on a path which connects the origin to $C$ and has length at most $L$", their $C$ being the unit circle; the question repeats the phrase of Cohen's theorem, and the posers report that Mac Lane answered it in the negative. The site's commentary states Cohen's theorem in the words of the site's question, a path in the set that connects $0$ to the unit circle, which is true only with the starting point exempt. The second insertion is the corpus's own correction. It excludes exactly the polynomials of degree $0$, at which no path can meet the conclusion. At degree $1$, $P(z)=1-z/\omega$ with $\lvert\omega\rvert=1$, the radius from $0$ to $\omega$ has $\lvert P\rvert<1$ except at $0$ and length $1$, and at every degree $n\ge1$ Cohen's theorem gives a path, so no other degree fails this way. The defect is already in the posers' question, which states the exemption for Cohen's path but not again in the question, and says nothing of degree $0$. The one result about the site's wording is the trivial answer above, the corpus's own observation, published nowhere else; it is credited here and counts for nothing. Mac Lane's theorem answers the corrected Statement no, and the problem's label and standing judge the corrected Statement.