Necessary condition for a function to be one-to-one

Question

A complex-valued function $f(z)$, is called one-to-one if $f(z_1) = f(z_2)$ implies $z_1 = z_2$.

The following statement is from Saff's Complex Analysis book.

If $f$ is analytic at $z_{0}$ and $f'\left(z_{0}\right) \neq 0$, then there is an open disk $D$ centered at $z_{0}$ such that $f$ is one-to-one on $D$.

The proof is as follows which I don't quite understand how the first implication is obtained and how the second implication leads to function being one-to-one. Any help in understanding this is much appreciated.

Answer 1

• $\frac{|f’(z_0)|}{2}$ is a positive number. So, you can take this as your “$\epsilon$” in the definition of $f’$ being continuous at $z_0$ (which it certainly is, since $f’$ is even better: analytic). This gives you the open disk $D$ centered at $z_0$.
• Consider any two points $z_1,z_2\in D$. The inequality in particular tells us that if $z_1\neq z_2$ then $|f(z_1)-f(z_2)|\geq |z_1-z_2|\cdot\frac{|f’(z_0)|}{2}>0$, since we have a product of two positive numbers, and hence $f(z_1)\neq f(z_2)$. This is the very definition of $f$ being injective (one-to-one) on $D$.

Another way of phrasing the injectivity argument from the inequality is to consider any two points $z_1,z_2\in D$ such that $f(z_1)=f(z_2)$. The inequality then implies $|z_1-z_2|\frac{|f’(z_0)|}{2}\leq |f(z_1)-f(z_2)|=0$, and since $|f’(z_0)|/2$ is non-zero, we can divide by it to get $|z_1-z_2|\leq 0$, and hence equality so $z_1=z_2$.