Why it is sufficient to show $|f'(z)-1|<1$?

According to an article entitled "On the Univalency of Certain Analytic Functions" by Wang et al. (2006), we have to show that $|f'(z)-1|<1$ in order to find the radius of univalency for the class $Q(\alpha,\beta,\gamma)$. Note that $Q(\alpha,\beta,\gamma)$ denotes the class of functions of the form $$f(z)=z+a_{2}z^{2}+\cdots$$ which are analytic in open unit disk, $D=\{z:|z|<1\}$, and satisfy the condition

$$\mathfrak{Re} \left\{\frac{\alpha f(z)}{z}+\beta f'(z)\right\}>\gamma \qquad (\alpha, \beta >0;\ 0 \leq \gamma<\alpha+ \beta\leq 1;\ z\in D)$$

Why it is sufficient to show that $|f'(z)-1|<1$?

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–  ˈjuː.zɚ79365 Jun 11 '13 at 14:42
Let $g(z)=f(z)-z$. The condition $|f'(z)-1|<1$ translates into $|g'(z)|<1$. Given distinct points $z,\zeta\in D$, integrate along the line segment between them to obtain $|g(z)-g(\zeta)|<|z-\zeta|$. Hence $f(z)\ne f(\zeta)$.