A holomorphic and injective function has nonzero derivative

This post proves that if f:U\to V is a function that is holomorphic (analytic) and injective, then f'(z)\neq 0 for all z in U. The condition of having nonzero derivative is equivalent to the condition of conformal (preserves angles). Hence, this result can be stated as “A holomoprhic and injective function is conformal.”

(Proof modified from Stein-Shakarchi Complex Analysis)

We prove by contradiction. Suppose to the contrary f'(z_0)=0 for some z_0\in D. Using Taylor series, \displaystyle f(z)=f(z_0)+f'(z_0)(z-z_0)+\frac{f''(z_0)}{2!}(z-z_0)^2+\dots

Since f'(z_0)=0, \displaystyle f(z)-f(z_0)=a(z-z_0)^k+G(z) for all z near z_0, with a\neq 0, k\geq 2 and G(z)=(z-z_0)^{k+1}H(z) where H is analytic.

For sufficiently small w\neq 0, we write \displaystyle f(z)-f(z_0)-w=F(z)+G(z), where F(z)=a(z-z_0)^k-w.

Since |G(z)|<|F(z)| on a small circle centered at z_0, and F has at least two zeroes inside that circle, Rouche’s theorem implies that f(z)-f(z_0)-w has at least two zeroes there.

Since the zeroes of a non-constant holomorphic function are isolated, f'(z)\neq 0 for all z\neq z_0 but sufficiently close to z_0.

Let z_1, z_2 be the two roots of f(z)-f(z_0)-w. Note that since w\neq 0, z_1\neq z_0, z_2\neq z_0. If z_1=z_2, then f(z)-f(z_0)-w=(z-z_1)^2h(z) for some analytic function h. This means f'(z_0)=0 which is a contradiction.

Thus z_1\neq z_2, which implies that f is not injective.

Author: mathtuition88


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.