On the univalence of polyanalytic functions

Theorem 1. Let $F(z) = \sum_{k=0}^{\alpha-1} \overline{z}^k A_k(z)$ be a poly-analytic function of order $\alpha$ on U, where $A_k$ are analytic such that $A_k(0) = 0$ , $A'_k(0) = 1$ and $|A_k| \leq M$ for all k. Then there is a constant $0 < \rho_1 < 1$ so that F is univalent in $|z| < \rho_1$ . In specific, $\rho_1$ satisfies $$1 - M\left(\frac{\rho_1(2 - \rho_1)}{(1 - \rho_1)^2} + \sum_{k=1}^{\alpha - 1} \frac{\rho_1^k(1 + k - \rho_1)}{(1 - \rho_1)^2}\right) = 0,$$ and $F(U_{\rho_1})$ contains a disk $U_{R_1}$ , where $R_1 = \rho_1 - \rho_1^2(\frac{1-\rho_1^{\alpha-1}}{1-\rho_1}) - M \sum_{k=0}^{\alpha-1} \frac{\rho_1^{k+2}}{1-\rho_1}$ . Proof. Fix $0 < \rho < 1$ and choose $z_1, z_2$ with $z_1 \neq z_2$ , $|z_1| < \rho$ and $|z_2| < \rho$ . We first note that $F_z = \sum_{k=0}^{\alpha-1} \overline{z}^k A_k'$ , $F_{\overline{z}}(z) = \sum_{k=0}^{\alpha-1} k \overline{z}^{k-1} A_k'$ , which implies $F_z(0) = A_0'(0) = 1$ , $F_{\overline{z}}(0) = A_1(0) = 0$ . We write each $A_k(z) = \sum_{n=1}^{\infty} a_{n,k} z^n$ . It follows that on the line segment $[z_1, z_2]$ we have $$\begin{split} &|F(z_{1}) - F(z_{2})| \\ &= \int\limits_{[z_{1},z_{2}]} F_{z}(z)dz + F_{\overline{z}}(z)d\overline{z} \\ &= \left| \int\limits_{[z_{1},z_{2}]} (F_{z}(0)dz + F_{\overline{z}}(0)d\overline{z}) + \int\limits_{[z_{1},z_{2}]} (F_{z}(z) - F_{z}(0)) dz + (F_{\overline{z}}(z) - F_{\overline{z}}(0)) d\overline{z} \right| \\ &= \left| \int\limits_{[z_{1},z_{2}]} dz + \int\limits_{[z_{1},z_{2}]} (A'_{0}(z) - A'_{0}(0))dz + \int\limits_{[z_{1},z_{2}]} \sum_{k=1}^{\alpha-1} k\overline{z}^{k-1} A_{k} d\overline{z} \right| \\ &\geq |z_{2} - z_{1}| - \int\limits_{[z_{1},z_{2}]} \sum_{n=2}^{\infty} n|a_{n,0}|\rho^{n-1} - \int\limits_{[z_{1},z_{2}]} \sum_{k=1}^{\alpha-1} |\overline{z}^{k}| \left( |A'_{k}| + k \left| \frac{A_{k}}{z} \right| \right) |dz| \\ &\geq |z_{2} - z_{1}| \left( 1 - \sum_{n=1}^{\infty} (n+1)|a_{n,0}|\rho^{n} - \sum_{k=1}^{\alpha-1} \rho^{k} \sum_{n=1}^{\infty} (n|a_{n,k}| + ka_{n,k}) \rho^{n-1} \right) \\ &\geq |z_{2} - z_{1}| \left( 1 - M \left( \frac{\rho}{(1-\rho)^{2}} + \frac{\rho}{1-\rho} \right) - M \sum_{k=1}^{\alpha-1} \rho^{k} \left( \frac{1}{(1-\rho)^{2}} + \frac{k}{1-\rho} \right) \right) \\ &= |z_{2} - z_{1}| \left( 1 - M \left( \frac{\rho(2-\rho)}{(1-\rho)^{2}} + \sum_{k=1}^{\alpha-1} \frac{\rho^{k}(1+k-\rho)}{(1-\rho)^{2}} \right) \right). \end{split}$$ Clearly there is a $\rho$ so that $|F(z_1) - F(z_2)| > 0$ . Let $\rho_1$ be the largest such $\rho$ . In other words, choose $\rho_1 > 0$ so that $$1 - M\left(\frac{\rho_1(2 - \rho_1)}{(1 - \rho_1)^2} + \sum_{k=1}^{\alpha - 1} \frac{\rho_1^k(1 + k - \rho_1)}{(1 - \rho_1)^2}\right) = 0.$$ For $|z|=\rho_1$ , $$|F(z)| = \left| \sum_{k=0}^{\alpha - 1} \overline{z}^k A_k(z) \right| = \left| a_{1,0} \ z + \sum_{k=1}^{\alpha - 1} a_{1,k} |z|^2 \overline{z}^{k-1} + \sum_{k=0}^{\alpha - 1} \overline{z}^k \sum_{k=0}^{\infty} a_{n,k} z^n \right|$$ $$\geq \rho_1 - \sum_{k=2}^{\alpha - 1} \rho_1^k - M \sum_{k=0}^{\alpha - 1} \sum_{k=0}^{\infty} \rho_1^{n+k}$$ $$= \rho_1 - \rho_1^2 \left( \frac{1 - \rho_1^{\alpha - 1}}{1 - \rho_1} \right) - M \sum_{k=0}^{\alpha - 1} \frac{\rho_1^{k+2}}{1 - \rho_1}.$$ As a corollary, we conclude a version of Landau's theorem for bi-analytic functions:

Corollary 1. Let $F(z) = \overline{z}A(z) + B(z)$ be a Bi-analytic function of U, where A and B are analytic such that A(0) = B(0) = 0, A'(0) = B'(0) = 1 and |A| and |B| are both bounded by M. Then there is a constant $0 < \rho_1 < 1$ so that F is univalent in $|z| < \rho_1$ . In specific, $\rho_1$ satisfies $$1 - 2M\left(\frac{2\rho_1 - \rho_1^2}{(1 - \rho_1)^2}\right) = 0,$$ that is $$\rho_1 = \frac{2M}{2M+1} \left( 1 + \sqrt{\frac{2M+1}{2M}} + \frac{1}{2M} \right).$$ Moreover, $F(U_{\rho_1})$ contains a disk $U_{R_1}$ , where $$R_1 = \rho_1 - \rho_1^2 - M \frac{\rho_1^3 + \rho_1^2}{1 - \rho_1}$$

Theorem 2. Let $F(z) = \sum_{k=0}^{\alpha-1} \overline{z}^k A_k(z)$ be a Poly-analytic function of order $\alpha$ , where $A_k$ are analytic mappings for $k = 0, 1, ...\alpha - 1$ , such that $f_k(z) = A_0(z) + \overline{A_k(z)}$ preserves orientation for each k. Suppose that $A_0$ is univalent and normalized by $A_0(0) = 0$ , $A_0'(0) = 1$ and $F(U) \subset U$ . Then $$M(F,r) < 1$$ if $|z| < r_0$ , where $r_0$ is the root of the polynomial $r^{\alpha} + r^{\alpha-1} + ... + r^3 + 3r - 1 = 0$ . We can take $r_0 \approx 0.318$ .

Corollary 2. Let $F(z) = \overline{z}A(z) + B(z)$ be a Bi-analytic function such that $f(z) = \overline{A}(z) + B(z)$ preserves the orientation. Suppose that B is univalent and normalized by B(0) = 0, B'(0) = 1 and $F(U) \subset U$ . Then $$M(F,r) < 1$$ if $|z| < \frac{1}{3}$ . The bound is sharp and is attained by suitable rotation of the Koebe function $A(z) = \frac{z}{(1-z)^2}$ . Recently, Abu Muhanna has shown the following lemma for analytic functions [3], which gives the radius under which the majorant function is bounded by the distance to the boundary of the image of an analytic function.

Lemma 1. Let $A(z) = \sum_{n=0}^{\infty} a_n z^n$ be analytic on U and suppose that A(U) misses at least two points then $$M(A) \leq dist(A(0), \partial A(U)),$$ for $$|z| \le e^{-\pi} = 4.3214 \times 10^{-2}$$ . We generalize the above lemma to poly-analytic functions under certain conditions on $A_k$ .

Theorem 3. Let $F(z) = \sum_{k=0}^{\alpha-1} \overline{z}^k A_k(z)$ be a Poly-analytic function of order $\alpha$ , where $A_k$ are analytic mappings for $k = 0, 1, ...\alpha - 1$ , such that $f_k(z) = A_0(z) + \overline{A_k(z)}$ preserves orientation for each k. Suppose that $A_0(z)$ misses at least two points and $A_0(0) = a_0$ , $A_k(0) = 0$ , for $k = 1, ..., \alpha - 1$ . Then if $|z| < e^{-\pi}$ , $$M(F,r) \le \frac{1-r^{\alpha}}{1-r}d(a_0,\partial A_0(U)).$$

Corollary 3. Let $F(z) = \overline{z}A(z) + B(z)$ be a Bi-analytic function such that $f(z) = B(z) + \overline{A(z)}$ preserves the orientation, where $B(0) = b_0$ , A(0) = 0, and suppose that B(z) misses at least two points. Then if $|z| < e^{-\pi}$ , we have $$M(F,r) \leq (1+r)d(b_0,\partial B(U)).$$

Theorem 4. Let $F(z) = \sum_{k=1}^{\alpha-1} \overline{z}^k A_k(z)$ be a Poly-analytic functions such that each $A_k(z)$ is a starlike analytic function and $|A_k(z)| \leq M(r)$ , for each $k = 1, ... \alpha - 1$ , 0 < r < 1. Let L(r) denote the arclength of the curve $C_r$ , where $C_r$ denote the image of the circle |z| = r < 1 under the function w = F(z). Then $$L(r) \le \frac{2\pi M(r)r}{1-r} \left[ \frac{(1+r)\left((\alpha-1)r^{\alpha} - \alpha r^{\alpha-1} + 1\right)}{(1-r)^2} - 1 + r^{\alpha-1} \right].$$

Theorem 5. Let $f(z) = \overline{z}A(z)$ be a bi-analytic function defined on the unit disk U such that A(z)/z is starlike univalent analytic. Let $M_p(r, f)$ denote the moment of order $p, p \geq 0$ . Then, $$M_p(r,f) \ge 2\pi \frac{r^{3p+6}}{3p+6}$$ . Equality holds if and only if $f(z) = \overline{z}z^2$

Theorem 6. Let $f(z) = \overline{z}A(z) + B(z)$ be a bi-analytic function defined on the unit disk U such that A(z)/z is starlike univalent analytic and $Re(\overline{z}A'\overline{B'}) \geq 0$ . Then, the minimal area is given by $$\int_{U_r} J_F dA \ge \frac{\pi r^6}{3},$$ where $U_r = \{z : |z| \le r\}.$

Theorem 7. Let $F(z) = \overline{z}A(z)$ be a Bi-analytic function, where $A \in H(U)$ , with A(0) = 0 and A'(0) = 1. A is starlike if and only if $\Phi(z) = zF(z)$ is starlike.