MATHEMATICS

Full Member of the Bulgarian Academy of Sciences Lyubomir N. CHAKALOV

ON DOMAINS OF UNIVALENCE OF CERTAIN CLASSES OF ANALYTIC FUNCTIONS

(Presented by Academician I. M. Vinogradov, 1 III 1960)

If the derivative \(P'_n(z)\) of a polynomial \(P_n(z)\) of degree \(n>1\) does not vanish in the disk \(|z|<1\), then, by S. Kakeya’s theorem \((^{1})\), the polynomial \(P_n(z)\) is univalent in the concentric disk of radius \(\sin \frac{\pi}{n}\). We shall show how the radius of univalence in Kakeya’s theorem can be increased by taking into account the distribution of the zeros of \(P'_n(z)\) outside the unit disk. The method we use is based on the following lemma.

Lemma. Let \(n>1\) be an integer. Consider \(n\) positive numbers
\(\alpha_1,\alpha_2,\ldots,\alpha_n\), whose sum \(S=\sum \alpha_k\) does not exceed \(\pi/2\), and \(n\) complex numbers \(u_1,u_2,\ldots,u_n\), whose moduli satisfy the inequalities

\[ |u_k|\leq \sin \alpha_k,\qquad k=1,2,\ldots,n. \tag{1} \]

Under these assumptions, the real part of the product \(\prod_k(1+u_k)\) is nonnegative:

\[ \operatorname{Re}\prod_k(1+u_k)\geq 0. \tag{2} \]

The equality sign in relation (2) occurs only when \(S=\pi/2\),
\(u_k=i\sin\alpha_k e^{i\alpha_k}\), \(k=1,2,\ldots,n\), or
\(u_k=-i\sin\alpha_k e^{i\alpha_k}\), \(k=1,2,\ldots,n\).

Proof. Put \(u_k=r_k e^{i\varphi_k}\), \(1+u_k=\rho_k e^{i\psi_k}\), where \(0<r_k\leq \sin\alpha_k\) and \(\rho_k>0\). We may assume that \(-\pi/2<\psi_k<\pi/2\), since \(\cos\psi_k>0\). Eliminating \(\rho_k\) from the equations

\[ 1+r_k\cos\varphi_k=\rho_k\cos\psi_k,\qquad r_k\sin\varphi_k=\rho_k\sin\psi_k, \]

we find

\[ \sin\psi_k=r_k\sin(\varphi_k-\psi_k), \]

\[ |\sin\psi_k|\leq r_k\leq \sin\alpha_k,\qquad |\psi_k|\leq \alpha_k, \tag{3} \]

\[ \left|\sum_k \psi_k\right|\leq \sum \alpha_k\leq \frac{\pi}{2}. \]

The assertions of the lemma follow from these relations.

Theorem 1. Let \(Q(z)\) be a polynomial in \(z\) of degree \(n>1\), whose roots \(z_1,z_2,\ldots,z_n\) are nonzero, and let \(|z_k|=r_k\). Then the polynomial of degree \(n+1\)

\[ P(z)=\int Q(z)\,dz \]

univalent in the disk \(|z| \leqslant r\), where \(r\) denotes the positive root of the equation

\[ \sum_k \arcsin \frac{r}{r_k}=\frac{\pi}{2}. \]

Proof. Without loss of generality, we may assume that \(Q(0)=1\) and, consequently,

\[ Q(z)=\prod_k\left(1-\frac{z}{z_k}\right). \tag{4} \]

Choosing arbitrarily \(n\) positive numbers \(\alpha_1,\alpha_2,\ldots,\alpha_n\), subject to the condition \(\sum \alpha_k \leqslant \frac{\pi}{2}\), and applying the lemma to the product (4), we conclude that the real part of \(Q(z)\) is nonnegative if

\[ \frac{|z|}{r_k}\leqslant \sin \alpha_k,\qquad |z|\leqslant r_k\sin\alpha_k, \]

that is, if \(|z|\) does not exceed the least of the products

\[ r_k\sin\alpha_k,\qquad k=1,2,\ldots,n. \]

In order to choose the \(\alpha_k\) in the most advantageous way, let us note that one can prove the existence of \(n\) positive numbers \(\alpha'_1,\alpha'_2,\ldots,\alpha'_n\), subject to the inequality \(\sum \alpha'_k \leqslant \frac{\pi}{2}\), for which the least of the numbers

\[ r_1\sin\alpha'_1,\ r_2\sin\alpha'_2,\ldots,\ r_n\sin\alpha'_n \tag{5} \]

has the greatest possible value. We shall show that for this it is necessary and sufficient that the conditions

\[ \sum \alpha'_k=\frac{\pi}{2},\qquad r_1\sin\alpha'_1=r_2\sin\alpha'_2=\cdots=r_n\sin\alpha'_n \]

be satisfied.

Indeed, let the number \(\mu\)—the least of the numbers (5)—have the greatest possible value. If \(\sum \alpha'_k<\frac{\pi}{2}\), then one can determine a positive number \(\varepsilon\) so that the inequality \(\sum \alpha_k<\frac{\pi}{2}\) holds, where \(\alpha_k=\alpha'_k+\varepsilon\). In that case the least of the numbers \(r_k\sin\alpha_k\) will be greater than \(\mu\), which is impossible. On the other hand, if the numbers (5) are not equal to one another, then among them at least one, for example \(r_m\sin\alpha'_m\), is greater than \(\mu\). Putting \(\alpha_k=\alpha'_k+\varepsilon\) for \(k\ne m\) and \(\alpha_m=\alpha'_m-(n-1)\varepsilon\), we conclude that \(\sum \alpha_k=\sum \alpha'_k\) and that, for sufficiently small positive \(\varepsilon\), each of the numbers \(r_k\sin\alpha_k\) is greater than \(\mu\), i.e., we again arrive at a contradiction.

If we denote by \(r\) the common value of the numbers (5) and eliminate \(\alpha'_1,\alpha'_2,\ldots,\alpha'_n\) from the equations

\[ \sum \alpha'_k=\frac{\pi}{2},\qquad r_k\sin\alpha'_k=r\quad (k=1,2,\ldots,n), \]

then for \(r\) we obtain the equation

\[ \sum_k \arcsin \frac{r}{r_k}=\frac{\pi}{2}, \tag{6} \]

which has exactly one positive root, since its left-hand side is an increasing function of \(r\) when \(r\) varies from \(0\) to

smallest of the numbers \(r_k\). In the disk \(|z|<r\) we have \(\operatorname{Re} P'(z)>0\), whence, as is known, it follows that the function \(P(z)\) is univalent in \(|z|\leq r\). Thus the theorem is proved.

The following theorem gives more concrete indications concerning the value of the radius \(r\), depending on the distribution of the roots of \(P'(z)=Q(z)\).

Theorem 2. Denote by \(m\) a natural number \(<\dfrac{n+1}{2}\) and by \(R\) the number
\[ R=\sin \frac{\pi}{n+1}:\sin \frac{(n+1-2m)\pi}{(n-m)(2n+2)}. \]
If \(m\) of the roots of the polynomial
\[ Q(z)=\prod_{k=1}^{n}\left(1-\frac{z}{z_k}\right) \]
lie in the annulus \(1\leq |z|\leq R\), and the remaining \(n-m\) lie in the region \(|z|>R\), then the radius of univalence \(r\) of the polynomial
\[ P(z)=\int Q(z)\,dz, \]
defined from equation (6), will be greater than the radius
\[ \sin \frac{\pi}{n+1}, \]
which is given by Kakeya’s theorem.

Proof. Let \(r\) be the positive root of equation (6), where \(r_k=|z_k|\) and \(0<r_1\leq r_2\leq \cdots \leq r_n\). Write equation (6) in the form
\[ \sum_{1}^{m}\arcsin \frac{r}{r_k}+\sum_{m+1}^{n}\arcsin \frac{r}{r_k}=\frac{\pi}{2}. \]
In the first sum we substitute \(1\) for \(r_k\), and in the second, \(R\). Then we obtain the inequality
\[ m\arcsin r+(n-m)\arcsin \frac{r}{R}>\frac{\pi}{2}, \]
from which it follows that the positive root \(r'\) of the equation
\[ m\arcsin r'+(n-m)\arcsin \frac{r'}{R}=\frac{\pi}{2} \]
is smaller than \(r\). Substituting the value of \(R\), we easily find
\[ r'=\sin \frac{\pi}{n+1}. \]
Thus \(r>\sin \dfrac{\pi}{n+1}\), which proves the theorem.

Corollary. For \(m=1\) we have
\[ R=2\cos \frac{\pi}{2n+2}. \]
From Theorem 2 we conclude that if \(|z_1|=1\) and \(|z_k|>2\cos \dfrac{\pi}{2n+2}\) for \(k=2,3,\ldots,n\), then the polynomial
\[ P(z)=\int Q(z)\,dz \]
is univalent in some disk \(|z|\leq r\), where
\[ r>\sin \frac{\pi}{n+1}. \]

In conclusion we note that if the polynomials \(Q_1(z)\) and \(Q_2(z)\), of degrees \(p\) and \(q\), do not vanish at \(z=0\), then our method can be applied to the function
\[ P(z)=\int \frac{Q_1(z)}{Q_2(z)}\,dz, \]
i.e. one can establish its univalence in some disk with center \(z=0\). If
\[ Q_1(z)=\prod_{k=1}^{p}\left(1-\frac{z}{z_k}\right) \]
and
\[ Q_2(z)=\prod_{k=p+1}^{p+q}\left(1-\frac{z}{z_k}\right), \]
then it is easy to prove that in the disk \(|z|\leq r\), where \(r\) is the positive root of the equation
\[ \sum_{1}^{p+q}\arcsin \frac{r}{|z_k|}=\frac{\pi}{2}, \]
the real part of the function
\[ R(r)=\frac{Q_1(z)}{Q_2(z)} \]
is nonnegative, and the function \(P(z)\) is univalent.

Bulgarian
Academy of Sciences

Received
29 II 1960

References

1. S. Kakeya, Tôhoku Math. J., 11, 5 (1917).