1 Introduction

It is known that for every complex valued function f on a symmetric set \(\mathcal {G}\) \(\subset \mathbb {C}\) \((-1\mathcal {G}=\mathcal {G})\), there exists exactly one function \(f_{E}\) from the class \(\mathcal {F}_{E}(\mathcal {G})\) of even functions and exactly one function \(f_{O}\) from the class \(\mathcal {F} _{O}(\mathcal {G})\) of odd functions on G, such that

$$\begin{aligned} f=f_{E}+f_{O}. \end{aligned}$$

Moreover,

$$\begin{aligned} f_{E}\left( z\right) =\frac{1}{2}\left[ f(z)+f(-z)\right] , f_{O}\left( z\right) =\frac{1}{2}\left[ f(z)-f(-z)\right] ,z\in \mathcal {G}. \end{aligned}$$

In view of the uniqueness of this partition, the functions \(f_{E}\) and \(f_{O} \) are called the even and odd part of f, respectively.

This partition has interesting applications, see for instance the proof of the Borsuk sandwich theorem 1.4.1. in the paper [1] and see also [5]. Of course, not every complex function of a symmetric set \(\mathcal {G}\) \(\subset \mathbb {C}\) is the product of an even and an odd function. For example no non-zero even function can be represented by the product of an even and an odd function. We will show that a function decomposition into the product of two functional factors is possible, if we define other special kinds of functional symmetry and a family of selected functions \(f:\mathcal {G\rightarrow }\mathbb {C}\).

2 Definitions and basic properties

Let \(\mathcal {G}\) \(\subset \mathbb {C}\) be a symmetric set such that \(0\in \mathcal {G}\). Then, by \(\mathcal {B}\) let us denote the set of all continuous functions \(f:\mathcal {G}\rightarrow \mathbb {C}\) normalized by \(f(0)=1\) and with \({\text {Re}}f(z)>0,z\in \mathcal {G}\). Of course, then \(f(z)\ne 0\) for \(z\in \mathcal {G}\).

Now we give the definition of the mentioned functional symmetries as well as the proposed names and designations.

Definition 1

The function \(f\in \mathcal {B}\) will be called \(\left( 1\right) \)-power symmetric and \(\left( -1\right) \)-power symmetric, if it satisfies the following conditions

$$\begin{aligned} f(-z)&=\left( f(z)\right) ^{1},z\in \mathcal {G},\\ f(-z)&=\left( f(z)\right) ^{-1},z\in \mathcal {G}, \end{aligned}$$

respectively.

For \(j=1,-1\) the family of all (j)-power symmetric functions \(f\in \mathcal {B}\) will be denoted by \(\mathcal {B}_{j}\). Obviously, the constant function \(f=1\) belongs to both families \(\mathcal {B}_{1}\),\(\mathcal {B}_{-1}\) and there does not exist any other such function. Let us show that the families \(\mathcal {B}_{j},j=1,-1\) are non-trivial.

Example

1. Let the symmetric set \(\mathcal {G}\) be the horizontal strip

$$\begin{aligned} \mathcal {G=}\left\{ z\in \mathbb {C},z=x+iy,x\in \mathbb {R}\wedge y\in \left( -\frac{\pi }{2},\frac{\pi }{2}\right) \right\} . \end{aligned}$$

Then the function \(f(z)=exp(z),z\in \mathcal {G},\) belongs to \(\mathcal {B}_{1}.\)

Indeed, the function f belongs to \(\mathcal {B},\) because f is continuous, \(f(0)=1\) and \({\text {Re}}\exp (z)=e^{x}\cos y>0\) for \(z\in \mathcal {G}.\) On the other hand

$$\begin{aligned} f(-z)=\exp {(-z)=}\frac{1}{f\left( z\right) },z\in \mathcal {G}. \end{aligned}$$

Hence \(f\in \mathcal {B}_{-1}.\) Of course, \(f\notin \mathcal {B}_{-1},\) if we replace the horizontal strip \(\mathcal {G}\) by the plane \(\mathbb {C}\).

2. Let the symmetric set \(\mathcal {G}\) be the vertical strip

$$\begin{aligned} \mathcal {G}=\left\{ z\in \mathbb {C},z=x+iy,x\in (-\frac{\pi }{2},\frac{\pi }{2}){\wedge }y\in \mathbb {R}\right\} . \end{aligned}$$

Then the function \(f(z)=\cos \left( z\right) ,z\in \mathcal {G}\) belongs to \(\mathcal {B}_{1}.\) Of course

\(f\in \mathcal {B},\) because f is continuous, normalized by \(f(0)=1\) and for \(z\in \mathcal {G}\) satisfies the condition

$$\begin{aligned} {\text {Re}}f(z)={\text {Re}}\,\cos {(z)}=Re\,\frac{e^{iz}+e^{-iz} }{2}=\frac{(e^{-y}+e^{y})\cos {x}}{2}>0. \end{aligned}$$

For the relation \(f\in \mathcal {B}_{1}\) it suffices to observe that \(\cos \left( -z\right) =\cos \left( z\right) ,z\in \mathcal {G}\).

Note that \(f\notin \mathcal {B}_{1}\) if we replace the set \(\mathcal {G}\) by the plane \(\mathbb {C}\).

3. Similarly, the function \(f(z)=\cosh \left( z\right) =\frac{e^{z}+e^{-z}}{2},\) reduced to the above horizontal strip \(\mathcal {G}\), belongs to the family \(\mathcal {B}_{1}.\)

4. Let us observe, that the functions \(f(z)=\sin \left( z\right) \) and \(f(z)=\sinh \left( z\right) \) reduced to a symmetric set do not belong to \(\mathcal {B}_{1}\) or \(\mathcal {B}_{-1},\) because \(f(0)\ne 1\).

5. Let the symmetric set \(\mathcal {G}\) be the open unit disc \(\mathcal {U}=\left\{ z\in \mathbb {C}:\left| z\right| <1\right\} .\) Then the function

$$\begin{aligned} f(z)=\frac{1}{1+z^{2}},z\in \mathcal {U} \end{aligned}$$

belongs to \(\mathcal {B}_{1}.\)

Indeed, the function f belongs to \(\mathcal {B},\) because f is continuous, \(f(0)=1\) and \({\text {Re}}f(z)>\frac{1}{2}.\) Now it suffices to observe that

$$\begin{aligned} f(-z)=f(z),z\in \mathcal {U}. \end{aligned}$$

From now, we will always take the unit disc \(\mathcal {U}\) as the symmetric set \(\mathcal {G}\).

In the paper, for a precise definition of the mentioned decomposition of a function into the product of a (1)-power symmetric and a \((-1)\)-power symmetric function, we use the following very useful properties of the continuous unique main branch of the square root

$$\begin{aligned} \sqrt{\cdot }:\mathbb {C}\ \backslash (-\infty ,0]\rightarrow \mathbb {C},\sqrt{1}=1 \end{aligned}$$

(the symbol \(\sqrt{\cdot }\) in the whole paper means the above main branch of the square root).

Lemma 2.1

Let

$$\begin{aligned} \pi ^{+}=\left\{ z\in \mathbb {C}:Re\,z>0\right\} . \end{aligned}$$

For \(z,w\in \) \(\pi ^{+}\) and the above main branch of the square root the following equalities hold:

$$\begin{aligned} (i)\; \sqrt{zw}&=\sqrt{z}\cdot \sqrt{w},\\ (ii)\;\sqrt{\frac{1}{z}}&=\frac{1}{\sqrt{z}}. \end{aligned}$$

Note that Lemma 2.1. is a direct consequence of the following result [6, Chap. 5] for the continuous unique main branch of the logarithm

$$\begin{aligned} Log:\mathbb {C}\ \backslash (-\infty ,0]\rightarrow \mathbb {C},Log1=0. \end{aligned}$$

Proposition 1

For \(z,w\in \) \(\pi ^{+}\) and the above main branch of the logarithm, the following equalities hold:

$$\begin{aligned} Log\,(wz)= & {} Log\,{w}+Log\,{z},\\ Log\,(\frac{w}{z})= & {} Log\,{w}-Log\,{z}. \end{aligned}$$

Now, we give two theorems about the invariance of (j)-power symmetry \(j=1,-1\) with respect some functional operations.

We start with the case of operations of one function argument. Of course if \(f\in \mathcal {B}\), then the power \(f^{0}=1\) belongs to both families \(\mathcal {B}_{1},\mathcal {B}_{-1}.\)

On the other hand we have the following.

Theorem 2.1

Let n be an integer, \(f\in \mathcal {B}\), its power \(f^{n}\) belongs to \(\mathcal {B}\) and \(\sqrt{\cdot }\) be the main branch of the square root. Then the following implications are true:

$$\begin{aligned} (i)\;f&\in \mathcal {B}_{1}\Rightarrow f^{n}\in \mathcal {B}_{1},\\ (ii)\;f&\in \mathcal {B}_{-1}\Rightarrow f^{n}\in \mathcal {B}_{-1},\\ iii)\;f&\in \mathcal {B}_{1}\Rightarrow \sqrt{f}\in \mathcal {B}_{1},\\ (iv)\; f&\in \mathcal {B}_{-1}\Rightarrow \sqrt{f}\in \mathcal {B}_{-1}. \end{aligned}$$

Proof

(i) We take an \(n\in \mathbb {Z}\backslash \{0\}\), then for \(f\in \mathcal {B}_{1}\) and \(z\in \mathcal {U}\) we have:

$$\begin{aligned} f^{n}(-z)=[f(-z)]^{n}=[f(z)]^{n}=f^{n}(z). \end{aligned}$$

This gives the thesis \(\left( i\right) \). Similarly, for \(n\in \mathbb {Z}\backslash \{0\}\), \(f\in \mathcal {B}_{-1}\) and \(z\in \mathcal {U}\) we have

$$\begin{aligned} f^{n}(-z)=[f(-z)]^{n}=[\frac{1}{f(z)}]^{n}=\frac{1}{f^{n}(z)}. \end{aligned}$$

This gives the thesis \(\left( ii\right) .\)

(iii) Let \(f\in \mathcal {B}_{1}\) (of course \(\sqrt{f}\in \mathcal {B)}\). Then from the definition of the (1)-power symmetric function, we have for \(z\in \mathcal {U}\):

$$\begin{aligned} \sqrt{f}(-z)=\sqrt{f(-z)}=\sqrt{f(z)}=\sqrt{f}(z), \end{aligned}$$

hence \(\sqrt{f}\in \mathcal {B}_{1}.\)

(iv) Now let \(f\in \mathcal {B}_{-1}\), then from the definition of the \((-1) \)-power symmetric function and Lemma 2.1. we have successively for \(z\in \mathcal {U}\):

$$\begin{aligned} \sqrt{f}(-z)=\sqrt{f(-z)}=\sqrt{\frac{1}{f(z)}}=\frac{\sqrt{1}}{\sqrt{f(z)} }=\frac{1}{\sqrt{f(z)}}=\frac{1}{\sqrt{f}}(z), \end{aligned}$$

which proves that \(\sqrt{f}\in \mathcal {B}_{-1}\). \(\square \)

The next theorem concerns the invariance of the (j)-power symmetry \(j=1,-1\) with respect to some functional operations of two function arguments.

Theorem 2.2

Let \(f,g\in \mathcal {B},f\cdot g\in \) \(\mathcal {B}\) and \(\frac{f}{g}\in \) \(\mathcal {B}\). Then the following implications are true:

$$\begin{aligned} (i)\,f,g&\in \mathcal {B}_{1}\Rightarrow f\cdot g\in \mathcal {B}_{1},\\ (ii)\,f,g&\in \mathcal {B}_{-1}\Rightarrow f\cdot g\in \mathcal {B}_{-1},\\ (iii)\,f,g&\in \mathcal {B}_{1}\Rightarrow \frac{f}{g}\in \mathcal {B}_{1},\\ (iv)\,f,g&\in \mathcal {B}_{-1}\Rightarrow \frac{f}{g}\in \mathcal {B}_{-1}. \end{aligned}$$

Proof

(i) Let f, g \(\in \mathcal {B}_{1}\). Using the definition of the (1)-power symmetric function we show (i) as follows:

$$\begin{aligned} (f\cdot g)(-z)=f(-z)\cdot g(-z)=f(z)\cdot g(z)=(f\cdot g)(z),z\in \mathcal {U}. \end{aligned}$$

Thus \(f\cdot g\) is (1)-power symmetric.

(ii) Let \(f,g\in \) \(\mathcal {B}_{-1}\), then using the definition of the \((-1\))-power symmetric function we show similarly:

$$\begin{aligned} (f\cdot g)(-z)=f(-z)\cdot g(-z)=\frac{1}{f(z)}\cdot \frac{1}{g(z)}=\frac{1}{f(z)\cdot g(z)}=\frac{1}{(f\cdot g)}(z). \end{aligned}$$

Hence \(f\cdot g\) is \((-1)\)-power symmetric.

For the proof of parts (iii) and (iv) it suffices to save \(\frac{f}{g}=f\cdot \frac{1}{g}\) and use the above proved theses (i), (ii) and Theorem 2.1. (i), (ii),  for \(n=-1.\) \(\square \)

Now, we give the main theorem of the paper. It concerns the decomposition of a function into (1)- and \((-1)\)-power symmetric factors.

Theorem 2.3

Let \(f\in \mathcal {B}\). Then:

  1. (i)

    The functions

    $$\begin{aligned} f_{1}(z)&=\sqrt{f(z)\cdot f(-z)},z\in \mathcal {U},\\ f_{-1}(z)&=\sqrt{f(z)\cdot [f(-z)]^{-1}},\,z\in \mathcal {U}, \end{aligned}$$

    with the main branch of the square root,  are (1)- and \((-1)\)-power symmetric, respectively.

  2. (ii)

    The above function f can be represented as the product of the functions \(f_{1}\) and \(\ f_{-1}\) , i.e.

    $$\begin{aligned} f(z)=f_{1}(z)\cdot f_{-1}(z),\qquad z\in \mathcal {U}. \end{aligned}$$
  3. (iii)

    This decomposition of a function f is unique. It means that, if there exist g \(\in \mathcal {B}_{1}\) and \(h\in \) \(\mathcal {B} _{-1}\) such that, \(f(z)=g(z)\cdot h(z)\) for \(z\in \mathcal {U},\) then \(f_{1}=g\) and \(f_{-1}=h.\)

Proof

Firstly, we show that the functions \(f_{1},f_{-1}\) are well defined. Indeed, since \(Re\,f(-z)>0,\) for \(z\in U,\) we also have \(Re\,\frac{1}{f(-z)}>0,z\in U.\) Then

$$\begin{aligned} Arg\left( f\,(z)\cdot f(-z)\right) \in (-\pi ,\pi ) \end{aligned}$$

and

$$\begin{aligned} Arg\,\left( f(z)\cdot [f(-z)]^{-1}\right) \in (-\pi ,\pi ). \end{aligned}$$

Then the functions are well defined, because \(\sqrt{\cdot }\) means the main branch of the square root.

Now we show that \(f_{1}\) is (1)-power symmetric and \(f_{-1}\) is \((-1)\)-power symmetric. Let us note that \(f_{1}(0)=1=f_{-1}(0)\) and \(f_{1},f_{-1} :\mathcal {U}\rightarrow \pi ^{+}.\) Obviously the functions \(f_{1}\) and \(f_{-1}\) are continuous, because f is continuous, the product and quotient of continuous functions are continuous, the main branch of the square root is a continuous. From the form of \(f_{1},\) we have for \(z\in \mathcal {U}\):

$$\begin{aligned} f_{1}(-z)=\sqrt{f(-z)\cdot f(z)}=\sqrt{f(z)\cdot f(-z)}=f_{1}(z). \end{aligned}$$

Thus \(f_{1}\) is (1)-power symmetric. From the definition of \(f_{-1}\) and Lemma 2.1, we show that \(f_{-1}\) is \((-1)\)-power symmetric. This follows from the following equalities for \(z\in \mathcal {U}\):

$$\begin{aligned} \begin{aligned} f_{-1}(-z)&=\sqrt{f(-z)\cdot \frac{1}{f(z)}}=\sqrt{f(-z)}\cdot \frac{1}{\sqrt{f(z)}}=\frac{1}{\frac{1}{\sqrt{f(-z)}}\cdot \sqrt{f(z)}}\\&=\frac{1}{\sqrt{\frac{1}{f(-z)}}\cdot \sqrt{f(z)}}=\frac{1}{\sqrt{f(z)\cdot \frac{1}{f(-z)}}}=\frac{1}{f_{-1}(z)}. \end{aligned} \end{aligned}$$

This completes the proof of (i).

Next we show (ii). In view of thesis (i) and Lemma 2.1, we can write for \(z\in \mathcal {U}\) the product \(f_{1}\cdot \) \(f_{-1}\) of the functions \(f_{1},\) \(f_{-1},\) step by step:

$$\begin{aligned} f_{1}(z)\cdot f_{-1}(z)= & {} \sqrt{f(z)\cdot f(-z)}\cdot \sqrt{f(z)\cdot \frac{1}{f(-z)}}\\= & {} \sqrt{f(z)}\cdot \sqrt{f(-z)}\cdot \sqrt{f(z)}\cdot \frac{1}{\sqrt{f(-z)} }=(\sqrt{f(z)})^{2}=f(z). \end{aligned}$$

Now we will prove (iii),  i.e., that the decomposition

$$\begin{aligned} f(z)=f_{1}(z)\cdot f_{-1}(z),z\in \mathcal {U} \end{aligned}$$

is unique.

Suppose that: \(f(z)=g(z)\cdot h(z),z\in \mathcal {U},\) where \(g\in \mathcal {B}_{1}\), \(h\in \mathcal {B}_{-1}\). Then

$$\begin{aligned} \begin{aligned} f_{1}(z)&=\sqrt{f(z)\cdot f(-z)}=\sqrt{f(z)}\cdot \sqrt{f(-z)} =\sqrt{g(z)h(z)}\cdot \sqrt{g(-z)h(-z)}\\&=\sqrt{g(z)}\cdot \sqrt{h(z)}\cdot \sqrt{g(-z)}\cdot \sqrt{h(-z)}. \end{aligned} \end{aligned}$$

Thus, using the fact that \(g\in \mathcal {B}_{1},\) \(h\in \mathcal {B}_{-1}\) and Lemma 2.1, we get for \(z\in \mathcal {U}:\)

$$\begin{aligned} f_{1}(z)&=\sqrt{g(z)}\cdot \sqrt{h(z)}\cdot \sqrt{g(z)}\cdot \sqrt{\frac{1}{h(z)}}=(\sqrt{g(z)})^{2}\cdot \sqrt{h(z)}\cdot \frac{1}{\sqrt{h(z)}}\\&=(\sqrt{g(z)})^{2}=g(z). \end{aligned}$$

Similarly for \(z\in \mathcal {U}\):

$$\begin{aligned} \begin{aligned} f_{-1}(z)&=\sqrt{f(z)\cdot \frac{1}{f(-z)}}=\sqrt{f(z)}\cdot \sqrt{\frac{1}{f(-z)}}=\sqrt{f(z)}\cdot \frac{1}{\sqrt{f(-z)}}\\&=\sqrt{g(z)h(z)}\cdot \frac{1}{\sqrt{g(-z)h(-z)}}=\sqrt{g(z)}\cdot \sqrt{h(z)}\cdot \frac{1}{\sqrt{g(-z)}}\cdot \frac{1}{\sqrt{h(-z)}}. \end{aligned} \end{aligned}$$

Using once again Lemma 2.1 and the fact that \(g\in \mathcal {B}_{1} ,h\in \mathcal {B}_{-1},\) we get for \(z\in \mathcal {U}:\)

$$\begin{aligned} \begin{aligned} f_{-1}(z)&=\sqrt{g(z)}\cdot \sqrt{h(z)}\cdot \frac{1}{\sqrt{g(z)}}\cdot \frac{1}{\sqrt{\frac{1}{h(z)}}}=\sqrt{h(z)}\cdot \sqrt{\frac{1}{\frac{1}{h(z)} }}=\sqrt{h(z)}\cdot \sqrt{h(z)}\\&=(\sqrt{h(z)})^{2}=h(z). \end{aligned} \end{aligned}$$

This gives (iii) and the proof is complete. \(\square \)

By the uniqueness of the above decomposition, we will call the function \(f_{1}\) and \(f_{-1}\) the (1)- and \((-1)\)-power symmetric parts of the function f.

Now, we turn to examples involving the form of the decomposition of various functions \(f\in \mathcal {B}\) into (1)-power symmetric \(f_{1}\) and \((-1)\)-power symmetric \(f_{-1}\) factors. We will focus on examples of decompositions of functions from Caratheodory’s class of holomorphic functions \(f:\mathcal {U}\rightarrow \mathbb {C},f(0)=1\) with \({\text {Re}}f\left( z\right) >0,z\in \mathcal {U}.\) Caratheodory’s class of functions plays an important role in geometric function theory (see for instance [2]).

Example

1. Let us consider the function

$$\begin{aligned} {f(z)=\frac{1+z}{1-z},\ z\in }\mathcal {U}{\ .} \end{aligned}$$

The function f belongs to \(\mathcal {B}\), because it is continuous, f\((0)=1\) and \({f(U)=\pi ^{+}}\). Putting \(-z\) in place of z,  we obtain:

$$\begin{aligned} f\left( -z\right) =\frac{1-z}{1+z}=\frac{1}{f\left( z\right) }. \end{aligned}$$

Thus f is a \((-1)\)-power symmetric function. Hence, by the uniqueness of the decomposition of the function f in Theorem 2.3., we have \(f_{_{1}}(z)=1,\) and \(\ \ f_{_{-1}}\left( z\right) =f(z),\) for \(z\in \mathcal {U}\) .

2. Let us consider the function

$$\begin{aligned} f{(z)=\frac{1+z^{2}}{1-z^{2}},\ z\in }\mathcal {U}{.} \end{aligned}$$

We see that f is a superposition of the previous function with the square function \(w(z)=z^{2},z\in \mathcal {U}\), hence f \(\in \)\(\mathcal {B}\). Putting \(\left( -z\right) \) in place of\(\ \ z\), we obtain that \(f\left( -z\right) =f(z)\) for \(z\in \mathcal {U}.\) Thus f is a (1)-power symmetric function. Hence, by the uniqueness of the decomposition of the function f in Theorem 2.3, we obtain that \(f_{1}(z)=f\left( z\right) \) and \(f_{-1}(z)=1.\)

3. Let us consider the function

$$\begin{aligned} {f(z)=\frac{1}{1-z},\ z\in }\mathcal {U}. \end{aligned}$$

Of course, f is continuous, normalized by f\((0)=1\) and \({f(U)}\subset \pi ^{+}\). Then the function f belongs to the family \(\mathcal {B}.\) Putting \(\left( -z\right) \) in place of\(\ z,\) we obtain:

$$\begin{aligned} f\left( -z\right) =\frac{1}{1+z},z\in \mathcal {U}. \end{aligned}$$

We can see that \(f\left( -z\right) \ne f(z)\) and \(f\left( -z\right) \ne \frac{1}{f(z)}.\) Thus f is neither (1)-power symmetric nor \((-1)\)-power symmetric. Therefore we cannot use the previous method of decomposition. Below we determine directly the power symmetric parts \(f_{1}\) and \(f_{-1}\) of the function f:

$$\begin{aligned} f_{1}(z)= & {} \sqrt{f(z)\cdot f(-z)}=\sqrt{\frac{1}{1-z}\cdot \frac{1}{1+z}} =\sqrt{\frac{1}{1-z^{2}}}=\frac{1}{\sqrt{1-z^{2}}},z\in \mathcal {U},\\ f_{-1}(z)= & {} \sqrt{f(z)\cdot \frac{1}{f(-z)}}=\sqrt{\frac{1+z}{1-z}} ,z\in \mathcal {U}. \end{aligned}$$

3 Applications

Now we show the potential possibility of using (1)- and \((-1)\)- power symmetric functions to solve certain functional equations.

Let us therefore consider the equations:

$$\begin{aligned} f(z)\cdot f(-z)= & {} F(z),z\in \mathcal {U}, \end{aligned}$$
(1)
$$\begin{aligned} \frac{f(z)}{f(-z)}= & {} F(z),z\in \mathcal {U}, \end{aligned}$$
(2)

where \(F\in \mathcal {B}\) is a given function and \(f\in \mathcal {B}\) is an unknown function.

These equations belong to the collection of the following type of functional equations

$$\begin{aligned} G(z,f(z),f(g(z)))=0,z\in \mathcal {U}, \end{aligned}$$
(3)

where G, g are given functions and \(f\in \mathcal {B}\) is the wanted function (see Kuczma’s monograph [4]). This follows from the observation that Eq. (1) has the form

$$\begin{aligned} F(z)-f(z)\cdot f(g(z))=0, \end{aligned}$$

where \(g(z)=-z\), \(F\in \) \(\mathcal {B}\) is a fixed function and Eq. (2) has the form

$$\begin{aligned} F(z)-\frac{f(z)}{f(g(z))}=0, \end{aligned}$$

respectively.

Let us add that Eq. (1) corresponds to the Schröder-type equation, i.e., equation [4] of the form

$$\begin{aligned} f(g(z))=s\cdot f(z),z\in \mathcal {U}, \end{aligned}$$

where \(s\in \mathbb {C}\) and \(g(z)=-z\).

We will use Theorem 2.3. to solve Eqs. (1) and (2). We will prove the following:

Theorem 3.1

Equation (1) has a solution \(f\in \mathcal {B},\) if and only if \(F\in \) \(\mathcal {B}_{1}.\)

Proof

“\(\Longrightarrow \)” Let \(f\in \mathcal {B}\) be a solution of Eq. (1). Then by replacing z by \(\left( -z\right) ,\) we obtain:

$$\begin{aligned} f(-z).f(z)=F(-z),z\in \mathcal {U}. \end{aligned}$$

From the above and from (1) we can see that

$$\begin{aligned} F(-z)=F(z),z\in \mathcal {U}. \end{aligned}$$

Hence, \(F\in \) \(\mathcal {B}_{1}.\)

“\(\Longleftarrow \)” Let \(F\in \) \(\mathcal {B}_{1}.\) Then, applying decomposition (ii) from Theorem 2.3. to Eq. (1), we have for \(z\in \mathcal {U}\):

$$\begin{aligned} f_{1}(z)\cdot f_{-1}(z)\cdot f_{1}(-z)\cdot f_{-1}(-z)=F(z). \end{aligned}$$

Next, using part \(\left( i\right) \) of Theorem 2.3., we have further:

$$\begin{aligned} f_{1}(z)\cdot f_{-1}(z)\cdot f_{1}(z)\cdot \frac{1}{f_{-1}(z)}=F(z),z\in \mathcal {U}. \end{aligned}$$

Therefore:

$$\begin{aligned} f_{1}(z)=\sqrt{F(z)},z\in \mathcal {U}, \end{aligned}$$

(Let us recall that \(\sqrt{\cdot }\) means the unique continuous main branch \(\sqrt{\cdot }:\mathbb {C}\ \backslash (-\infty ,0]\rightarrow \) \(\mathbb {C} ,\sqrt{1}=1,\) of the square root).

We check that the function \(f(z)=\sqrt{F(z)}\cdot g(z)\), \(z\in \mathcal {U}\), with a \((-1)\)-power symmetric function g,  is a solution of Eq. (1). Indeed, since \(F\in \) \(\mathcal {B}_{1}\) and \(g\in \mathcal {B}_{-1}\), we obtain for \(z\in \mathcal {U}\):

$$\begin{aligned} f(z)\cdot f(-z)=\sqrt{F(z)}\cdot g(z)\cdot \sqrt{F(z)}\cdot \frac{1}{g(z)}=(\sqrt{F(z)})^{2}=F(z). \end{aligned}$$

This ends the proof. \(\square \)

Corollary

If \(F\in \) \(\mathcal {B}_{1},\) then every solution f of Eq. (1) has the form:

$$\begin{aligned} f(z)=\sqrt{F(z)}\cdot g(z),z\in \mathcal {U} \end{aligned}$$

with the main branch of the function \(\sqrt{\cdot }\) and with a \((-1)\)-power symmetric function g.

Theorem 3.2

Equation (2) has a solution \(f\in \mathcal {B},\) if and only if \(F\in \) \(\mathcal {B}_{-1}.\)

Proof

“\(\Longrightarrow ''\)Let \(f\in \mathcal {B}\) be a solution of Eq. (2). Then, putting \(\left( -z\right) \) in place of z, we obtain:

$$\begin{aligned} \frac{f(-z)}{f(z)}=F(-z),z\in \mathcal {U}. \end{aligned}$$

Multiplying this equation and Eq. (1) we obtain for \(z\in \mathcal {U}\):

$$\begin{aligned} \frac{f(z)}{f(-z)}\cdot \frac{f(-z)}{f(z)}=F(z)\cdot F(-z). \end{aligned}$$

Hence, we get that

$$\begin{aligned} F(-z)=\frac{1}{F(z)},z\in \mathcal {U}. \end{aligned}$$

Thus \(F\in \) \(\mathcal {B}_{-1}.\)

“\(\Longleftarrow ''\) Let \(F\in \) \(\mathcal {B}_{-1}.\) Then applying decomposition (ii) from Theorem 2.3 to Eq. (2), we have for \(z\in \mathcal {U}\):

$$\begin{aligned} \frac{f_{1}(z)\cdot f_{-1}(z)}{f_{1}(-z)\cdot f_{-1}(-z)}=F(z). \end{aligned}$$

Next, using part \(\left( i\right) \) of Theorem 2.3. we have further:

$$\begin{aligned} \frac{f_{1}(z)}{f_{1}(z)}\cdot f_{-1}(z)\cdot f_{-1}(z)=F(z),z\in \mathcal {U}. \end{aligned}$$

Hence

$$\begin{aligned} f_{-1}(z)=\sqrt{F(z)},\qquad z\in \mathcal {U}, \end{aligned}$$

where \(\sqrt{\cdot }\) means the main branch of the square root.

Now, we check that the function

$$\begin{aligned} f(z)=\sqrt{F(z)}\cdot g(z),z\in \mathcal {U}, \end{aligned}$$

with a \((-1)\)-power symmetric function g, is a solution of Eq. (2). To this aim, we observe that

$$\begin{aligned} \frac{f(z)}{f(-z)}=\frac{\sqrt{F(z)}\cdot g(z)}{\sqrt{F(-z)}\cdot g(-z)} ,z\in \mathcal {U}. \end{aligned}$$

Since \(F\in \) \(\mathcal {B}_{-1}\) and \(g\in \) \(\mathcal {B}_{1,}\)using Lemma 2.1. we obtain for \(z\in \mathcal {U}\)

$$\begin{aligned} \frac{f(z)}{f(-z)}=\frac{\sqrt{F(z)}\cdot g(z)}{\frac{1}{\sqrt{F(z)}}\cdot g(z)}=(\sqrt{F(z)})^{2}=F(z). \end{aligned}$$

This ends the proof. \(\square \)

Corollary

If \(F\in \) \(\mathcal {B}_{-1}\) then every solution of Eq. (2) has the form:

$$\begin{aligned} f(z)=\sqrt{F(z)}\cdot g(z),z\in \mathcal {U} \end{aligned}$$

with the main branch of the square root and a (1)-power symmetric function g.

Now we will show the possibility of using (j)-power-symmetric functions,  \(j=1,-1\), to investigate the properties of functions defined by geometrical conditions.

We start with the definition of starlikeness of order \(\alpha \in \left[ 0,1\right) .\)

Definition 2

We say that a holomorphic function \(g:\mathcal {U}\rightarrow \mathbb {C}\) , \(g\left( 0\right) =0,\) \(g^{\prime }\left( 0\right) =1\) belongs to the family \(\mathcal {S}^{*}\left( \alpha \right) ,\alpha \in \left[ 0,1\right) ,\) of starlike mappings of order \(\alpha ,\) if

$$\begin{aligned} {\text {Re}}\frac{zg^{\prime }\left( z\right) }{g(z)}>\alpha ,z\in \mathcal {U}. \end{aligned}$$

The following property of \(g\in \mathcal {S}^{*}\left( \alpha \right) \) is known (see [2, thm 2.3.4]).

Lemma 3.1

A function g belongs to \(\mathcal {S}^{*}\left( \alpha \right) ,\) if only if

$$\begin{aligned} G\left( z\right) =z\left( \frac{g\left( z\right) }{z}\right) ^{\frac{1}{1-\alpha }}\in \mathcal {S}^{*}. \end{aligned}$$

Here we use the continuous unique main branch \(\left( \cdot \right) ^{\frac{1}{1-\alpha }}:\mathbb {C}\ \backslash (-\infty ,0]\rightarrow \mathbb {C},\) \((1)^{^{\frac{1}{1-\alpha }}}=1,\) of the power function.

The mentioned application of functional power symmetries relates to a family \(K_{\mathcal {S}}\) (see [3]).

Definition 3

The holomorphic function \(f:\mathcal {U}\rightarrow \mathbb {C},f\left( 0\right) =0,f^{\prime }\left( 0\right) =1,\) belongs to the family \(\mathcal {K}_{\mathcal {S}}\) if there exists a function \(g\in \mathcal {S}^{*}\left( \frac{1}{2}\right) \) such that

$$\begin{aligned} {\text {Re}}\frac{z^{2}f^{\prime }\left( z\right) }{g(z)g(-z)} <0,z\in \mathcal {U}. \end{aligned}$$
(1)

We prove the following theorem

Theorem 3.3

For every \(f\in \mathcal {K}_{\mathcal {S}},\) there exsists \(g\in \mathcal {S} ^{*}\left( \frac{1}{2}\right) ,\) such that

$$\begin{aligned} {\text {Re}}\frac{f^{\prime }\left( z\right) }{\left( \left( \frac{g(z)}{z}\right) _{1}\right) ^{2}}>0,z\in \mathcal {U}, \end{aligned}$$

where the function \(\left( \frac{g(z)}{z}\right) _{1}\), is the \(\left( 1\right) \)-power symmetric part of the function \(\frac{g(z)}{z}.\)

Proof

Since \(g\in \mathcal {S}^{*}\left( \frac{1}{2}\right) ,\) we have from Lemma 3.1 that the function

$$\begin{aligned} G(z)=\frac{g^{2}(z)}{z},z\in \mathcal {U}\setminus \left\{ 0\right\} ,G(0)=1 \end{aligned}$$

belongs to the family \(\mathcal {S}^{*}\) of starlike univalent functions. Hence (see for instance [2, Thm. 2.3.5]) the function

$$\begin{aligned} G^{*}(z)=\sqrt{\frac{G(z)}{z}},G^{*}(0)=1,z\in \mathcal {U} \end{aligned}$$

with the main branch of the square root has positive real part in \(\mathcal {U}\) and also

$$\begin{aligned} G^{*}(z)=\frac{g(z)}{z},z\in \mathcal {U}. \end{aligned}$$

Consequently, the function \(\frac{g(z)}{z}\) belongs to the family \(\mathcal {B}\) and we can compute its \(\left( 1\right) \)-power symmetric part. Using the definition of the \(\left( 1\right) \)-power symmetric part of functions, we have:

$$\begin{aligned} G_{1}^{*}(z)=\sqrt{\frac{g(z)}{z}\cdot \frac{g(-z)}{-z}},z\in \mathcal {U}. \end{aligned}$$

Hence

$$\begin{aligned} \left( G_{1}^{*}(z)\right) ^{2}=\frac{g(z)}{z}\cdot \frac{g(-z)}{-z} ,z\in \mathcal {U}. \end{aligned}$$

Inserting this equality into the defining inequality (1) we obtain the thesis of Theorem 3.3. \(\square \)