Abstract Generated abstract
This paper studies univalence criteria for a finite class of meromorphic functions expressed as sums of principal parts with poles away from the origin. Under the assumption that the arguments of the ratios of coefficients to the corresponding pole powers are constant, the author proves that such functions are univalent in a closed disk centered at the origin whose radius is determined by the distance to the nearest pole and by the pole order. The proof uses a lemma giving a sharp disk in which the real part of a product of principal power factors remains positive, then applies this positivity to an integral representation of divided differences. The paper also notes that analogous conclusions may extend to broader rational meromorphic forms under suitable parameter conditions.
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PAVEL TODOROV
ON THE UNIVALENCE OF A CLASS OF MEROMORPHIC FUNCTIONS
(Presented by Academician M. A. Lavrent’ev on 28 X 1964)
As is known, an analytic function is meromorphic if, at finite distance, it has no singular points other than poles. Consider the class of meromorphic functions
\[ f(z)=\sum_{k=1}^{n}\frac{A_k}{(z-a_k)^m}, \tag{1} \]
where \(m \geqslant 1\) is an arbitrary natural number; \(A_k\) and \(a_k\), \(k=1,2,\ldots,n\), are two systems of constant complex numbers, none of which is equal to zero. Restrictions may be imposed on the poles \(a_k\) with respect to their position. Below we give some investigations connected with this problem. Under the condition that the argument of the ratio \(A_k/a_k^{m+1}\), \(k=1,2,\ldots,n\), is constant, we shall prove the following theorem.
Theorem. Each function of the form (1) is univalent in the closed circular domain
\[ |z|\leqslant \delta \sin \frac{\pi}{2(m+1)}, \tag{2} \]
where \(\delta\) is the distance of the nearest pole \(a_k\) from the origin.
For the proof of the theorem we shall use the following lemma.
Lemma. The real part of the product
\[ F(z_1,z_2,\ldots,z_n)=(1-z_1)^{m_1}(1-z_2)^{m_2}\ldots(1-z_n)^{m_n}, \quad n \geqslant 1, \tag{3} \]
where throughout the principal values of the power functions are taken and the real exponents \(|m_k|\geqslant 1\), \(k=1,2,\ldots,n\), are arbitrary, is always positive, i.e.
\[ \operatorname{Re} F(z_1,z_2,\ldots,z_n)>0, \tag{4} \]
when the complex numbers \(z_1,z_2,\ldots,z_n\) vary arbitrarily in the disk
\[ |z|<\sin \frac{\pi}{2n\max |m_k|}. \tag{5} \]
The number \(\sin \dfrac{\pi}{2n\max |m_k|}\) cannot be replaced by a larger number.
Proof. We note that the absolute value \(|\arg(1-z_k)|\) does not exceed the magnitude of the acute angle \(\dfrac{\pi}{2n|m_k|}\) between the tangent from the point \(z=1\) to the circle \(|z|=\sin \dfrac{\pi}{2n|m_k|}\) and the real axis, when \(z_k\) varies inside and on this circle, i.e.
\[ |\arg(1-z_k)|\leqslant \frac{\pi}{2n|m_k|}, \]
with equality attained only when \(z_k\) lies at the point of tangency. From
from the equality \(\arg F(z_1,z_2,\ldots,z_n)=\sum_{k=1}^n m_k\arg(1-z_k)\) it follows that \(|\arg F|\leqslant \dfrac{\pi}{2}\), and equality is obtained only for
\[ z_k=\sin\frac{\pi}{2n|m_k|} \left|\sin\frac{\pi}{2n|m_k|}\pm i\cos\frac{\pi}{2n|m_k|}\right|, \quad k=1,2,\ldots,n. \]
The minus sign corresponds to the second point of tangency. Only in this exceptional case is the product \(F(z_1,z_2,\ldots,z_n)\) a purely imaginary number. If one replaces the disk \(|z|<\sin\dfrac{\pi}{2n|m_k|}\) by the smallest one, i.e. by the disk
\[ |z|<\sin\frac{\pi}{2n\max |m_k|}, \]
then, under an arbitrary variation of the numbers \(z_1,z_2,\ldots,z_n\) in it, one always has \(\operatorname{Re}F>0\). It is clear that this disk is the largest with this property.
For the proof of the theorem let us note that if \(z_1\ne z_2\) are two points of the closed disk
\[ |z|\leqslant \delta\sin\frac{\pi}{2(m+1)}, \]
then the difference
\[ f(z_2)-f(z_1) = -m\sum_{k=1}^n A_k \int_{z_1}^{z_2}\frac{dt}{(z-a_k)^{m+1}}, \]
where the integration may be carried out along the straight segment \(z_1z_2\). When the real variable \(t\) increases from \(0\) to \(1\), the variable of integration
\[ z=(1-t)z_1+tz_2 \]
describes this segment and, consequently,
\[ -\frac{1}{m}\frac{f(z_2)-f(z_1)}{z_2-z_1} = (-1)^{m+1}\sum_{k=1}^n \frac{A_k}{a_k^{m+1}} \int_0^1 \frac{dt}{\left(1-\dfrac{z}{a_k}\right)^{m+1}}, \quad z=(1-t)z_1+tz_2. \tag{7} \]
Under the assumptions made, the real part of the integrand is positive in the interval \(0\leqslant t\leqslant 1\), or possibly is equal to zero at its endpoints, because, according to the lemma, for \(n=1\), \(m_1=m+1\), this is so when
\[ \left|\frac{z}{a_k}\right|\leqslant \frac{|z|}{\delta} \leqslant \sin\frac{\pi}{2(m+1)}. \]
Only in the case when \(|a_k|=\delta\) and at least one of the cases
\[ z_{1,2} = \delta\sin\frac{\pi}{2(m+1)} \left( \sin\frac{\pi}{2(m+1)} \pm i\cos\frac{\pi}{2(m+1)} \right), \]
holds, for \(t=0\) or \(1\), is the real part equal to zero. Consequently, the real parts of the integrals in (7) are positive. From the condition
\[ \arg\frac{A_k}{a_k^{m+1}}=\operatorname{const} \]
we obtain that all terms on the right-hand side of (7) lie on one side of the straight line passing through the origin (the one obtained by rotating the imaginary axis through the angle
\[ \arg\frac{A_k}{a_k^{m+1}} \]).
Consequently, the right-hand side of (7) is not equal to zero.
This proves the theorem.
The results obtained also hold for a broader class of functions
\[ f(z)=\sum_{k=1}^n \frac{A_k}{(z-a_k)^{m_k}(z-b_k)^{p_k}(z-c_k)^{q_k}\ldots} \]
under suitable conditions on the numbers
\[ A_k,\ a_k,\ b_k,\ c_k,\ldots,\ m_k,\ p_k,\ q_k,\ldots \]
Plovdiv
Bulgaria
Received
6 VIII 1964