Sufficient Conditons of Univalency for
Complex Functions in the Class C¹

1. Introduction

The following sufficient condition for univalency of analytic function in a convex domain is well known [2], [4], [5]:

Theorem A.

If $D$ is a convex domain in the complex plane $ℂ$, and if $f$ is an analytic function in $D$ such that

(1) $$\mathrm{Re}{f}^{\prime }\left(z\right)>0,\text{for all }z\in D,$$

then $f$ is univalent in $D$.

This was generalized in [1] and [3] as follows:

Theorem B.

If $f$ is analytic in a domain $D$ and if there exists an analytic function $g$ which is univalent and convex in $D$ (i.e. $g\left(D\right)is$ a convex domain) such that

(2) $$\mathrm{Re}\frac{f^{\prime }\left(z\right)}{g^{\prime }\left(z\right)}>0,\text{for all }z\in D,$$

then $f$ is univalent in $D$.

In this note we obtain sufficient conditions of univalency similar to (1) and (2) for complex functions in the class $C^1$. These conditions yield some simple criteria of homeomorphism in the complex plane.

2. Preliminaries

Let $D$ be a domain in $ℂ$ and let $f:D\to ℂ$, with $f\left(z\right)=u(x,y)+iv(x,y).$

We say that the function $f$ belongs to the class $C^1\left(D\right)$ if the real functions $u=\mathrm{Re}f$ and $v=\mathrm{Im}f$ of the real variables $x$ and $y$ have continuous first order partial derivatives in $D$.

The directional derivative of the function $f\in C^1\left(D\right)$ at the point $z\in D$, in the direction defined by the angle $\alpha \in [0,2\pi )$ is given by the well known formula

$$f_{\alpha }^{\prime }\left(z\right)=\frac{\partial f}{\partial z}+e^{-2i\alpha }\frac{\partial f}{\partial \overline{z}},$$

where

$$\frac{\partial }{\partial z}=\frac{1}{2}\left(\frac{\partial }{\partial x}-i\frac{\partial }{\partial y},\frac{\partial }{\partial \overline{z}}=\frac{1}{2}\frac{\partial }{\partial x}+i\frac{\partial }{\partial y}\right).$$

The Jacobian of the function $f\in C^1\left(D\right)$ is given by

$$J\left(f\right)=\left|\frac{\partial f}{\partial z}\right|-{\left|\frac{\partial f}{\partial \overline{z}}\right|}^2.$$

If $J\left(f\right)>0$ in $D$, then $f$ is a locally homeomorphism preserving the orientation in $D$.

For $f$ and $g$ belonging to $C^1\left(D\right)$, if we define

$$I(f,g)=\left|\begin{array}{c}\frac{\partial f}{\partial z}\frac{\partial f}{\partial \overline{z}}\\ \frac{\partial g}{\partial z}\frac{\partial g}{\partial \overline{z}}\end{array}\right|,$$

then $I(\overline{f},f)=J\left(f\right),$ and

(3) $${\left|I(f,\overline{g})\right|}^2-{\left|I(f,g)\right|}^2=J\left(f\right)\cdot J\left(g\right).$$

3. Main results

The following theorems provide sufficient conditions of homeomorphism similar to (1) and (2).

Theorem 1.

If $D$ is a convex domain in $ℂ$ and the function $f\in C^1\left(D\right)$ satisfies one of the following equivalent conditions:

(4) $$\mathrm{Re}{f}_{\alpha }^{\prime }\left(z\right)>0\text{for all }z\in D,\text{ and all }\alpha \in [0,2\pi )$$

(5) $$\mathrm{Re}\frac{\partial f}{\partial z}>\left|\frac{\partial f}{\partial \overline{z}}\right|,\text{for all }z\in D,$$

then $f$ is univalent in $D$.

Proof.

Let $z_1,z_2\in D$, $z\ne z_2$ and let $z\left(t\right)=z_1+t\left(z_2-z_1\right),t\in [0,1]$. Since $D$ is convex, $z\left(t\right)\in D$ for all $t\in [0,1]$. By integrating along the segment $[z_1,z_2],$ we obtain

$f\left(z_2\right)-f\left(z_1\right)$ $={\displaystyle \underset{0}{\overset{1}{\int }}}{\displaystyle \frac{d}{dt}}f\left[z\left(t\right)\right]𝑑t={\displaystyle \underset{0}{\overset{1}{\int }}}\left[{\displaystyle \frac{\partial f}{\partial z}}{\displaystyle \frac{dz}{dt}}+{\displaystyle \frac{\partial f}{\partial \overline{z}}}{\displaystyle \frac{d\overline{z}}{dt}}\right]𝑑t=$

$=\left(z_2-z_1\right){\displaystyle \underset{0}{\overset{1}{\int }}}\left|{\displaystyle \frac{\partial f}{\partial z}}+{\displaystyle \frac{{\overline{z}}_2-{\overline{z}}_1}{z_2-z_1}}{\displaystyle \frac{\partial f}{\partial \overline{z}}}\right|𝑑t=\left(z_2-z_1\right){\displaystyle \underset{0}{\overset{1}{\int }}}{f}_{\alpha }^{\prime }\left[z\left(t\right)\right]𝑑t,$

where $\alpha =\arg \left(z_2-z_1\right).$ By using (4) we obtain

$$\mathrm{Re}\frac{f\left(z_1\right)-f\left(z_2\right)}{z_1-z_2}=\underset{0}{\overset{1}{\int }}\mathrm{Re}{f}_{\alpha }^{\prime }\left[z\left(t\right)\right]𝑑t>0,$$

which shows that $f$ is univalent in $D$.

It is easy to check that the conditions (4) and (5) are equivalent.

We next generalize Theorem 1, by obtaining a sufficient condition of univalency similar to (2). ∎

Theorem 2.

Let $g\in C^1\left(D\right)$, where $D$ is a domain in $ℂ$. Suppose $g$ is univalent, convex and satisfies in $D$. If $f\in C^1\left(D\right)$ and

(6) $$\mathrm{Re}I(f,\overline{g})>\left|I(f,g)\right|,\text{for all }z\in D,$$

then $f$ is univalent and $J\left(f\right)>0$ in $D$.

Proof.

Let $△=g\left(D\right)$ and $\phi =g^{-1}$. Consider the function $h:△\to ℂ$ defined by

$$h\left(w\right)=I\left[\phi \left(w\right)\right],w\in △.$$

We have

${\displaystyle \frac{\partial h}{\partial w}}$ $={\displaystyle \frac{\partial f}{\partial z}}{\displaystyle \frac{\partial \phi }{\partial w}}+{\displaystyle \frac{\partial f}{\partial \overline{z}}}{\displaystyle \frac{\partial \overline{\phi }}{\partial w}}$

${\displaystyle \frac{\partial h}{\partial \overline{w}}}$ $={\displaystyle \frac{\partial f}{\partial z}}{\displaystyle \frac{\partial \phi }{\partial \overline{w}}}+{\displaystyle \frac{\partial f}{\partial \overline{z}}}{\displaystyle \frac{\partial \overline{\phi }}{\partial \overline{w}}}$

and

$$\frac{\partial \phi }{\partial w}=\frac{1}{J\left(g\right)}\frac{\partial \overline{g}}{\partial z},\frac{\partial \phi }{\partial \overline{w}}=-\frac{1}{J\left(g\right)}\frac{\partial g}{\partial \overline{z}}.$$

Hence

$$\frac{\partial h}{\partial w}=\frac{I(f,\overline{g})}{J\left(g\right)},\frac{\partial h}{\partial \overline{w}}=-\frac{I(f,g)}{J\left(g\right)}.$$

Since $J\left(g\right)>0,$ form (6) we obtain

$$\mathrm{Re}\frac{\partial h}{\partial w}>\left|\frac{\partial h}{\partial \overline{w}}\right|,w\in △.$$

Since $△$ is convex, by applying Theorem 1 we deduce that $h$ is univalent in $△$ and thus $f=h\circ g$ is univalent in $D$.

From (6) and (3) we easily deduce $J\left(f\right)>0,$ which completes the proof of Theorem 2 ∎

4. Particular Cases

• If the function $f$ in Theorem 2 is analytic in $D$, then the condition (6) becomes

  $$\mathrm{Re}\left|f^{\prime }\left(z\right)\frac{\partial \overline{g}}{\partial \overline{z}}\right|>\left|f^{\prime }\left(z\right)\frac{\partial g}{\partial \overline{z}}\right|,z\in D.$$

  If in addition $g$ is analytic in $D$, we obtain condition (2). If $D$ is convex and $g\left(z\right)\equiv z,$ we obtain condition (1).

• If the function $g$ in Theorem 2 is analytic in $D$, the condition (6) becomes

  $$\mathrm{Re}\left[\frac{\partial f}{\partial z}\overline{g^{\prime }\left(z\right)}\right]>\left|\frac{\partial f}{\partial \overline{z}}{g}^{\prime }\left(z\right)\right|,z\in D,$$

  or

  (7) $$\mathrm{Re}\frac{\frac{\partial f}{\partial z}}{g^{\prime }\left(z\right)}>\left|\frac{\frac{\partial f}{\partial \overline{z}}}{g^{\prime }\left(z\right)}\right|,z\in D.$$

  As an example, suppose $D$ is the unit disc $U,$ and $g\left(z\right)=z\left(1-z\right)$ or $g\left(z\right)=\log \left[\left(1+z\right)/\left(1-z\right)\right],z\in U$. Then we obtain respectively the following particular conditions of univalency for a function $f\in C^1\left(U\right):$

  $\mathrm{Re}\left[{(1-z)}^2{\displaystyle \frac{\partial f}{\partial z}}\right|$ $>\left|{\left(1-z\right)}^2{\displaystyle \frac{\partial f}{\partial \overline{z}}}\right|,z\in U,$

  $\mathrm{Re}\left[(1-z^2){\displaystyle \frac{\partial f}{\partial z}}\right|$ $>\left|\left(1-z^2\right){\displaystyle \frac{\partial f}{\partial \overline{z}}}\right|,z\in U.$

• If the function $f$ in Theorem 2 is of the form $f=F+\overline{G}$, where $F$ and $G$ are analytic in $D$, then condition (7) becomes

  (8) $$\mathrm{Re}\frac{F^{\prime }\left(z\right)}{g^{\prime }\left(z\right)}>\left|\frac{G^{\prime }\left(z\right)}{g^{\prime }\left(z\right)}\right|,z\in D.$$

  If in addition $D$ is convex and $g\left(z\right)\equiv z,$ we obtain the following condition of univalency for the function $f=F+G$

  $$\mathrm{Re}{F}^{\prime }\left(z\right)>\left|G^{\prime }\left(z\right)\right|,z\in D.$$

  (this also follows directly from Theorem 1).

  If $F=g$ AND for $z\in D$, then condition (8) holds. Hence we obtain the following condition of univalency

Theorem 3.

If the function $g$ is analytic univalent and convex in a domain $D$ and $G$ is analytic in $D$ such that for all $z\in D,$ then the function $f=g+G$ is univalent in $D$.

As an example, it is easy to check that the function $f\left(z\right)=z+e^{\overline{z}}$ is univalent in the halfplane .

• Consider $g\left(z\right)=z+\lambda z\overline{z}$ with $\left|\lambda \right|\le 1/2$. It is easy to show that $J\left(g\right)>0$ and $g$ is convex in $U$. By Theorem 1 we can show that $g$ is univalent in $U$. From Theorem 2 we obtain the following sufficient condition of univalency for a function $f\in C^1\left(U\right)$:

  $$\mathrm{Re}\left(\frac{\partial f}{\partial z}+\overline{\lambda }Df\left(z\right)\right)>\left|\frac{\partial f}{\partial \overline{z}}-\lambda Df\left(z\right)\right|,z\in U,$$

  where $\left|\lambda \right|\le 1/2$ and

  $$Df=z\frac{\partial f}{\partial z}-\overline{z}\frac{\partial f}{\partial \overline{z}}.$$

Received 10IIV.1979

Department of Mathematics,

“Babeş-Bolyai” University

3400 Cluj-Napoca