[CA] Why Injectivity Requires Non-zero Derivative at 1st Order for a Holomorphic Complex Function

cf. Greene Ch5 Problem #16 Question is shown by the title, assuming we have f holomorphic on some domain U. We can use Rouche's theorem to prove a stronger result, that is, zero derivative makes local injective impossible. Proof by contradiction: assume f'(P) = 0. We gonna prove that f has multiple preimages on any neighborhood of P from f(P), the process is the following: Take Taylor expansion at P as f is analytic on that, then move terms to consider f(z) - f(P) on the LHS. As f'(P) = 0, we have RHS starts from 2nd or even higher order (let's say, from k > 1) derivative. We take this starting term as g(z), and the sum of rest terms says h(z). Moving terms again, we have (f(z) - f(P)) - g(z) = h(z). Now considering g and h. When z is sufficiently close to P, comparing the order, |g(z)|>|h(z)|. So we have |(f(z) - f(P)) - g(z)| = |h(z)|<|g(z)|, by the weak version of Rouche, we know f(z) - f(P) has the same number of zeros with g(z), which has at least 2. Hence, f(z) = f(P) achieved twice around P counting multiplicity, which means f has multiple preimages, so not injective.