Univalent Functions with Half-Integral Coefficients

Abstract

B. Friedman discovered in his 1946 paper that the set of analytic univalent functions on the unit disk in the complex plane with integral Taylor coefficients consists of nine functions. In the present paper, we prove that the similar set obtained by replacing “integral” by “half-integral” consists of another 12 functions in addition to the nine. We also observe geometric properties of the 12 functions.

Appendix

In the present section, we collect several formulae which are useful in the Proof of Theorem 1.2. Let $f(z)=z+a_2z^2+a_3z^3+\cdots$ be in $\mathcal{A }$ and $1/f(z)=1/z+b_0+b_1z+b_2z^2+\cdots .$ We first note that the coefficients $b_n$ are computed in terms of $a_n$’s recursively by the formula

$$\begin{aligned} b_{n-1}=-a_{n+1}-\sum _{k=2}^n a_kb_{n-k},\quad n\ge 1. \end{aligned}$$

(5.1)

In particular, we have

$$\begin{aligned} b_0&= -a_2, \\ b_1&= -a_3+a_2^2, \\ b_2&= -a_4+2a_2a_3-a_2^3, \\ b_3&= -a_5+2a_2a_4+a_3^2-3a_2^2a_3+a_2^4, \\ b_4&= -a_6+2a_2a_5+2a_3a_4-3a_2^2a_4-3a_2a_3^2+4a_2^3a_3-a_2^5, \end{aligned}$$

and so on. The Grunsky coefficients $c_{j,k}$ of $f$ can be computed recursively by

$$\begin{aligned} c_{j,k}=\sum _{l=1}^{k-1}\frac{l}{k} a_{k-l}c_{j+1,l}-\sum _{m=1}^j a_{m+1}c_{j-m,k} -\frac{a_{j+k+1}}{k} \end{aligned}$$

for $j\ge 0$ and $k\ge 1$ (see [8] for details). Here, we set $a_1=1.$ It is easy to see that $c_{j,k}$ can be expressed as a polynomial in $a_2,\ldots ,a_{j+k+1}.$ We also note that $c_{j,k}=c_{k,j}.$ For convenience, we write down the coefficients $c_{j,k}$ for $1\le j\le k\le 3$ so that the reader can compute the Grunsky matrices of orders $2$ and $3:$

$$\begin{aligned} c_{1,1}&= -a_3+a_2^2, \\ c_{1,2}&= -a_4+2a_2a_3-a_2^3, \\ c_{1,3}&= -a_5+2a_2a_4+a_3^2-3a_2^2a_3+a_2^4, \\ c_{2,2}&= -a_5+2a_2a_4+\frac{3}{2}a_3^2-4a_2^2a_3+\frac{3}{2}a_2^4, \\ c_{2,3}&= -a_6+2a_2a_5+3a_3a_4-4a_2^2a_4-5a_2a_3^2+7a_2^3a_3-2a_2^5, \\ c_{3,3}&= -a_7+2a_2a_6+3a_3a_5-4a_2^2a_5+2a_4^2-12a_2a_3a_4 \\&\!\!\!+8a_2^3a_4-\frac{7}{3}a_3^3 +15a_2^2a_3^2-14a_2^4a_3+\frac{10}{3}a_2^6. \end{aligned}$$