Is an analytic map with a certain condition injective?

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    2017
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SUMMARY

The discussion centers on proving the injectivity of an analytic map \( f : \mathbb{D} \to \mathbb{C} \) defined by the power series \( f(z) = \sum_{n=0}^\infty a_n z^n \) for \( z \in \mathbb{D} \). The condition for injectivity is established as \( \sum_{n=2}^\infty n |a_n| < |a_1| \). This condition ensures that the map does not take the same value for different inputs within the unit disk, thereby confirming its injective nature.

PREREQUISITES
  • Understanding of analytic functions and their properties
  • Familiarity with power series and convergence
  • Knowledge of the concept of injectivity in complex analysis
  • Basic grasp of series manipulation and inequalities
NEXT STEPS
  • Study the properties of analytic functions in complex analysis
  • Learn about the implications of the Schwarz lemma on injective maps
  • Explore the role of power series in defining analytic functions
  • Investigate conditions for injectivity in higher-dimensional complex spaces
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Mathematicians, particularly those specializing in complex analysis, students studying analytic functions, and researchers interested in the properties of injective mappings in the complex plane.

Euge
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Here is this week's POTW:

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Consider an analytic map $f : \Bbb D \to \Bbb C$ such that $f(z) = \sum\limits_{n = 0}^\infty a_n z^n$ for all $z\in \Bbb D$. Prove that $f$ is injective, provided

$$\sum_{n = 2}^\infty n\lvert a_n\rvert < \lvert a_1\rvert.$$

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This week's problem was solved by Kokuhaku. You can read his solution below.
Note that $f'(z)-a_1 = \sum_{n=2}^\infty na_n$ for $z \in \mathbb{D}$, so we actually have $|f'(z)-a_1|<|a_1|$ for $z \in \mathbb{D}$ (use triangle inequality).

Suppose now that $z_1 \neq z_2$ (from $\mathbb{D}$). Then using above inequality and estimation lemma (M-L inequality) we have
\begin{align*}
|f(z_2)-f(z_1)| &= \Bigl|\int_{z_1}^{z_2} f'(z)\, dz\Bigr| = \Bigl|a_1 \int_{z_1}^{z_2} dz + \int_{z_1}^{z_2} (f'(z) - a_1) \, dz\Bigr| \\
&\ge |a_1(z_2-z_1)| - \Bigl|\int_{z_1}^{z_2} (f'(z) - a_1) \, dz\Bigr| > |a_1||z_2-z_1| -|z_2-z_1||a_1| = 0,
\end{align*}
so $f(z_1) \neq f(z_2)$, which gives us that $f$ is injective function.
 

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