Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $|f'(z)-1|<1$?

share|cite|improve this question
up vote 12 down vote accepted

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 $f(z)\ne f(\zeta)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.