UDC 517.535.6

MATHEMATICS

PAVEL TODOROV

ON THE POLES AND THE RADIUS OF CIRCULAR UNIVALENCE OF THE MEROMORPHIC CLASS \(\Phi\)

(Presented by Academician M. A. Lavrent’ev, 27 XI 1965)

In \((^1)\) we considered one of the subclasses of the meromorphic class \(\Phi\):

\[ f(z)=\sum_{k=1}^{n}\sum_{s=1}^{m_k}\frac{A_{ks}}{(z-a_k)^s}, \tag{1} \]

\[ a_k=\delta_k e^{i\varphi_k},\quad \delta_k>0,\quad m_k\geq 1 \text{—integers,} \]

\[ \arg A_{ks}-(s+1)(\pi+\arg a_k)\equiv \mathrm{const}\equiv \varphi \pmod{2\pi} \tag{2} \]

for all \(k,s\) with \(A_{ks}\ne 0\), and established the radius of a circle with center at \(z=0\) such that every function of the subclass under consideration is univalent in it. In the present study we solve the problem completely.

Theorem 1. Let \(R(\delta_k,m_k)\) be the greatest of the numbers \(R\) such that every function of the class \(\Phi\) is univalent in the circle \(|z|\leq R\). Then

\[ R(\delta_k,m_k)=\min_k \delta_k \sin \frac{\pi}{2(m_k+1)}. \tag{3} \]

The proof of Theorem 1 is based on two lemmas and two theorems.

Lemma 1 (a generalization of the lemma from \((^1)\)). Let \(p_1,p_2,\ldots,p_n\), \(n\geq 1\), be arbitrary real numbers, and suppose the powers \((1-z_k)^{p_k}\), \(k=1,2,\ldots,n\), have their principal value. Then the real part of the function

\[ P(z_1,\ldots,z_n)=(1-z_1)^{p_1}\cdots(1-z_n)^{p_n} \]

is positive for arbitrary complex numbers \(z_1,\ldots,z_n\) lying in the circle

\[ |z|<\sin\frac{\pi}{2p},\quad p=\sum_{k=1}^{n}|p_k| \]

when \(p>1\), and in the circle \(|z|<1\) when \(0<p\leq 1\). The numbers \(\sin(\pi/2p)\) for \(p>1\) and \(1\) for \(p\leq 1\) cannot be replaced by larger ones.

Proof. From the inequalities \(|\arg(1-z)|<\gamma\) for \(|z|<\sin\gamma\), \(0<\gamma_0\leq \pi/2\), it follows that

\[ |\arg P(z_1,\ldots,z_n)|<\gamma\sum_{k=1}^{n}|p_k|=p\gamma. \]

Putting here \(\gamma=\gamma_0=\pi/2p\) when \(p>1\), or \(\gamma=1\) when \(0<p\leq 1\), we obtain the first assertion of the lemma. To prove the unimprovability of these bounds for \(p>1\), set, for \(p_k>0\), \(z_k=\sin(\pi/2p)e^{-i\alpha}\), where \(\alpha=(\pi/2p)(p-1)\), and, for \(p_k<0\), \(z_k=\sin(\pi/2p)e^{i\alpha}\); it is easy to verify that in this case \(\arg P(z_1,\ldots,z_n)=\pi/2\). If \(p\leq 1\) and at least one of the numbers \(z_k\) is equal to 1, then either \(P(z_1,\ldots,z_n)=0\), or \(P\) has no meaning. Thus both circles in the lemma are the largest possible with the required properties.

Lemma 2. Let the function \(f(z)\) be holomorphic in the circle \(|z-z_0|<r_0\), univalent in the closed circle \(|z-z_0|\leq c\) \((<r_0)\), and let on its circumference \(|z-z_0|=c\) the derivative \(f'(z)\ne 0\). Then there exists at least one \(c'>c\) such that \(f(z)\) is univalent in the closed circle \(|z-z_0|\leq c\).

Proof. Otherwise, for any sequence \(\varepsilon_n \to 0\) as \(n \to \infty\) \((\varepsilon_n>0)\), there will be found a sequence of pairs of points \(z_1^{(n)}, z_2^{(n)}\) such that \(z_1^{(n)} \ne z_2^{(n)}\), \(|z_j^{(n)}-z_0|\le c+\varepsilon_n\) \((j=1,2)\), and \(f(z_2^{(n)})=f(z_1^{(n)})\).

Let \(z_j^{(n_k)}\) be a convergent subsequence, \(z_j^{(n_k)} \to z_j^*\). Clearly \(z_j^*\) lies in the disk \(|z-z_0|\le c\), and by the continuity of \(f(z)\) we have \(f(z_1^*)=f(z_2^*)\). Since \(f\) is univalent in the disk \(|z-z_0|\le c\), it follows that \(z_1^*=z_2^*\). Consequently, in every neighborhood of \(z_1^*\) there lies a pair of points \(z_1^{(n_k)}\) and \(z_2^{(n_k)}\) such that \(z_1^{(n_k)}\ne z_2^{(n_k)}\) and \(f(z_1^{(n_k)})=f(z_2^{(n_k)})\). Under the assumptions made about the univalence of \(f\), the point \(z_1^*\) cannot lie in the disk \(|z-z_0|<c\). However, the case \(|z-z_0|=c\) is impossible, since \(f'(z_1^*)\ne 0\), and therefore \(f(z)\) is univalent in a neighborhood of \(z_1^*\). The lemma is proved.

Theorem 2. The disk \(|z|\le R(\delta_k,m_k)\) (where \(R(\delta_k,m_k)\) is defined by equality (3)) is a domain of univalence for every function of the class \(\Phi\).

Proof. For \(z_1\ne z_2\) the equality holds

\[ \frac{-e^{-i\varphi}(f(z_2)-f(z_1))}{z_2-z_1} = \sum_{k=1}^{n}\sum_{s=1}^{m_k} \frac{s}{\delta_k^{s+1}}|A_{ks}| \int_{0}^{1} \frac{dt}{(1-z/a_k)^{s+1}}, \tag{4} \]

where \(z=(1-t)z_1+tz_2\), and it is assumed that the segment \(z_1z_2\) does not pass through a pole of \(f(z)\). According to Lemma 1, the real parts of the integrand functions are positive when

\[ \left|\frac{z}{a_k}\right| < \sin \frac{\pi}{2(m_k+1)} \le \sin \frac{\pi}{2(s+1)}, \qquad k=1,\ldots,n;\ s=1,\ldots,m_k . \tag{5} \]

Consequently, for \(|z_1|, |z_2| < \min_k \delta_k \sin \pi/2(m_k+1)\) (recall that \(\delta_k=|a_k|\)) the right-hand side of equality (4) is nonzero. Thus, for \(z_1\ne z_2\) we have \(f(z_1)\ne f(z_2)\), as was required to prove.

By the criterion given in Lemma 2, the disk \(|z|\le R\) will be the largest disk in which \(f(z)\) is univalent if on its circumference there is at least one point \(z_0\) such that \(f'(z_0)=0\). In the following theorem all functions of the class \(\Phi\) with this property are described.

Theorem 3. Let an arbitrary function of the class \(\Phi\) have a zero on the circumference of the disk of radius \(R\) with center at \(z=0\). Then \(f(z)\) has the form

\[ e^{-i\varphi}f(z)= \tag{6} \]

\[ = \sum_{\sigma=1}^{\tau}\sum_{k=1}^{\nu_\sigma} \frac{ d_{k\sigma}\exp\left\{ i(m_\sigma+1) \left( \pi+\varphi_{11} + \frac{\pi\varepsilon_{11}m_1}{2(m_1+1)} - \frac{\pi\varepsilon_{k\sigma}m_\sigma}{2(m_\sigma+1)} \right) \right\} }{ \left\{ z - R\cosec\frac{\pi}{2(m_\sigma+1)} \exp\left( i\varphi_{11} + \frac{i\pi\varepsilon_{11}m_1}{2(m_1+1)} - \frac{i\pi\varepsilon_{k\sigma}m_\sigma}{2(m_\sigma+1)} \right) \right\}^{m_\sigma} }, \]

where

\[ \sum_{\sigma=1}^{\tau}\sum_{k=1}^{\nu_\sigma} \varepsilon_{k\sigma}m_\sigma d_{k\sigma} \left\{ \frac{1}{R}\lg \frac{\pi}{2(m_\sigma+1)} \right\}^{m_\sigma+1} =0 . \tag{7} \]

Only for these functions is the closed disk \(|z|\le R\) the largest disk of univalence.

Proof. For the derivative of a function of the class \(\Phi\) we have

\[ -e^{-i\varphi}f'(z) = \sum_{k=1}^{n}\sum_{s=1}^{m_k} \frac{s}{\delta_k^{s+1}}|A_{ks}| \left(1-\frac{z}{a_k}\right)^{-s-1}. \tag{8} \]

I. First of all, if for some \(z\) with \(|z|=R\) we have \(f'(z)=0\), then \(A_{ks}=0\) for \(s<m_k,\ k=1,\ldots,n\). Indeed, from \(|z|\le |a_k|\sin \frac{\pi}{2(m_k+1)}<\)

\(< \delta_k \sin \dfrac{\pi}{2(s+1)}\) and \(|z|\le R\) it follows that
\(\operatorname{Re}\{(1-z_n/a_k)^{-s-1}\}>0\), i.e. \(f'(z)\ne0\) for \(|z|=R\), and \(A_{ks}\ne0\) with \(s<m_k\).

II. Further, if \(f'(z)\) vanishes on the circle \(|z|=R\), then

\[ \delta_k \sin \frac{\pi}{2(m_k+1)}=R \tag{9} \]

for all \(k\) (otherwise, among the nonnegative quantities \(\operatorname{Re}(1-z/a_k)^{-s-1}\) there would be a strictly positive one, and the equality \(f'(z)=0\) would be impossible for \(|z|=R\)).

III. From I and II we find the form of those functions of the class \(\Phi\) whose first derivative vanishes on the circle \(|z|=R\):

\[ f(z)=\sum_{\sigma=1}^{\tau}\sum_{k=1}^{\nu_\sigma} \frac{A_{k\sigma}}{(z-a_{k\sigma})^{m_\sigma}}, \tag{10} \]

where \(a_{k\sigma}\ne a_{\sigma k}\) for \(k\ne\sigma\), \(a_{k\sigma}=\delta_\sigma e^{i\varphi_{k\sigma}}\) for \(k=1,\ldots,\nu_\sigma\); \(\delta_\sigma<\delta_a\) and \(m_\sigma<m_a\) for \(\sigma<a\). The derivative of a function of the form (10) is equal to (we put \(A_{k\sigma}=d_{k\sigma}\exp(i\beta_{k\sigma})\))

\[ -e^{-i\varphi}f'(z)= \sum_{\sigma=1}^{\tau}\sum_{k=1}^{\nu_\sigma} m_\sigma d_{k\sigma}\left\{\delta_\sigma\left(1-\frac{z}{a_k}\right)\right\}^{-m_\sigma-1}. \tag{11} \]

According to Lemma 1, from the equality \(f'(\zeta)=0\), \(\zeta=Re^{i\varphi}\), it follows that

\[ \zeta=\delta_\sigma e^{i\varphi_{k\sigma}}\sin\frac{\pi}{2(m_\sigma+1)} \exp\left(\frac{i\pi m_\sigma \varepsilon_{k\sigma}}{2(m_\sigma+1)}\right), \quad k=1,\ldots,\nu_\sigma;\ \sigma=1,\ldots,\tau, \tag{12} \]

where \(\varepsilon_{k\sigma}=\pm1\) must not have a constant sign. From (9), (11), and (12) we obtain (7). Further, from (12) we have

\[ \varphi_{k\sigma}\equiv \varphi_{11}+ \frac{\pi\varepsilon_{11}m_1}{2(m_2+1)} - \frac{\pi\varepsilon_{k\sigma}m_\sigma}{2(m_\sigma+1)} \pmod{2\pi}. \tag{13} \]

Further,

\[ \zeta= R\exp\left(i\varphi_{11}+\frac{i\varepsilon_{11}\pi m_1}{2(m_1+1)}\right) = \frac12 a_{11}\left(1-\exp\left(-\frac{i\pi\varepsilon_{11}}{m_1+1}\right)\right). \]

From the condition that a function of the form (10) belongs to the class \(\Phi\), and from (13), we find

\[ \beta_{k\sigma}\equiv \varphi+(m_\sigma+1)\left( \pi+\varphi_{11}+ \frac{\pi\varepsilon_{11}m_1}{2(m_1+1)} - \frac{\pi\varepsilon_{k\sigma}m_\sigma}{2(m_\sigma+1)} \right) \pmod{2\pi}. \tag{14} \]

Substituting (9), (13), and (14) into (10), we obtain (6), as was required to prove.

If we put \(m_1=\max_\sigma m_\sigma\), then from (6) it immediately follows that \(\tau=1\), \(k=1,\ldots,\nu_1\), and (5) becomes \(\sum_{k=1}^{\nu_1}\varepsilon_{k1}d_{k1}=0\). Hence \(\sum_k' d_{k1}=\sum_k'' d_{k1}=d\), where \(\sum_k'\), \(\sum_k''\) denote sums over those \(k\) for which \(\varepsilon_{k1}=+1\), respectively \(\varepsilon_{k1}=-1\). Consequently, (6) turns into

\[ e^{-i\varphi}f(z)= \tag{6'} \]

\[ = \frac{ d\exp\bigl(i(m_1+1)(\pi+\varphi_{11})\bigr) }{ \left( z-R\cosec\frac{\pi}{2(m_1+1)}e^{i\varphi_{11}} \right)^{m_1} } + \frac{ d\exp\{i(m_1+1)(\pi+\varphi_{11})+i\pi\varepsilon_{11}m_1\} }{ \left( z+R\cosec\frac{\pi}{2(m_1+1)} e^{\,i\left(\varphi_{11}-\frac{\pi\varepsilon_{11}}{m_1+1}\right)} \right)^{m_1} }, \]

i.e. we have obtained those, among the functions considered in (1), for which the disk \(|z|\le R\) is the largest disk of univalence.

Theorem 1 follows from Theorem 2 and Theorem 3.

Received
27 XI 1965

CITED LITERATURE

1. Pavel Todorov, DAN, 162, No. 2 (1965).