Show that if \(f\) is injective, then \(f'\) is nowhere vanishing and thus \(f\) is conformal.
Solution
Contrapositive: if \(f'(z_0)=0\) at some point, then \(f\) fails injectivity in a neighborhood of \(z_0\). Without loss of generality, we can assume \(f(0) = 0\) after changes of coordinates in the domain/codomain, since this won’t affect injectivity or vanishing of derivatives. If \(z_0\) is a zero of \(f\) of some order \(n\), we can write \(f(z) = z^n g(z)\) with \(g(0) \neq 0\) and \(g\) nonvanishing in some neighborhood of zero. Choosing a branch of \(z\mapsto z^{1\over n}\), write \(g(z) = (h(z))^n\) for some \(h\), then \[ f(z) = z^n h(z)^n = (zh(z))^n .\] Write \(H(z) \da zh(z)\), then \(H'(z) = h(z) + zh'(z)\), so \(H'(0)\neq 0\). By the inverse function theorem, \(H\) is invertible on a small neighborhood of \(0\), making \(f\) \(n\)-to-one for some \(n\). Writing \(f(z) = \sum_{k\geq 0} c_k z^k\), we have \(c_0 = 0\) since \(f(0) = 0\) and \(c_1 = 0\) if \(f'(0) = 0\), making \(n\geq 2\), so \(f\) fails injectivity in this neighborhood.