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Abstract
A conjecture of Bombieri states that the coefficients of a normalized univalent function $f$ should satisfy $$ \liminf_{f\to K} \frac{n-{\rm Re\,}a_n}{m-{\rm Re\,}a_m} = \min_{t\in{\mathbb R}} \, \frac{n\sin t -\sin(nt)}{m\sin t -\sin(mt)}, $$ when $f$ approaches the Koebe function $K(z)=\frac{z}{(1-z)^2}$. Recently, Leung disproved this conjecture for $n=2$ and for all $m\geq3$ and, also, for $n=3$ and for all odd $m\geq5$. Complementing his work we disprove it for all $m>n\geq2$ which are simu
Results & Lemmas (8)
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Theorem 1.
Theorem 1. Let m > n ≥2 be integers such that either (a) both m and n are odd, or (b) both m and n are even, or (c) m is odd, n is even and…
Theorem 1. Let m > n ≥2 be integers such that either (a) both m and n are odd, or (b) both m and n are even, or (c) m is odd, n is even and n ≤m+1 2 . Then (6) is true. We have already observed that one can deduce the following corollary. Corollary. Let m > n ≥2 be integers such that either (a), (b) or (c) in
Theorem 1
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Theorem 1 holds. Then Bombieri’s conjecture for this pair of integers is false.
Theorem 1 holds. Then Bombieri’s conjecture for this pair of integers is false.
Theorem 1
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Theorem 1 will be proved mainly with the use of trigonometry, but also, in the case when the hypothesis (c) holds, we will employ…
Theorem 1 will be proved mainly with the use of trigonometry, but also, in the case when the hypothesis (c) holds, we will employ Dieudonn´e’s criterion for univalent polynomials. After carefully examining the relevant graphs for 2 ≤n ≤80 using the www.desmos.com/calculator software, one is lead to believe that the hypothe- sis (c) in Theorem 1 can be notably weakened in that the point (m, n) has to be below the straight line that joins the points (7, 6) and (17, 14). Thus, the following proposi
Lemma 2
Lemma 2 (Dieudonn´e’s criterion). The polynomial p(z) = z + a2z2 +... + anzn is univalent in D if and only if its associated polynomials…
Lemma 2 (Dieudonn´e’s criterion). The polynomial p(z) = z + a2z2 + . . . + anzn is univalent in D if and only if its associated polynomials q(z; t) = 1 + a2 sin(2t) sin t z + . . . + an sin(nt) sin t zn−1 have no zeros in D for any choice of the parameter t ∈[0, π]. We now prove a simple lemma for An(t) = n −sin(nt) sin t , which we defined in (2).
Lemma 3.
Lemma 3. For all t ∈R and n ≥2, we have An(t) ≥0 and An(2π −t) = An(t). Also, An vanishes only for t = 2ℓπ, ℓ∈Z, when n is even and only…
Lemma 3. For all t ∈R and n ≥2, we have An(t) ≥0 and An(2π −t) = An(t). Also, An vanishes only for t = 2ℓπ, ℓ∈Z, when n is even and only for t = ℓπ, ℓ∈Z, when n is odd.
Lemma 4.
Lemma 4. For all integers n ≥2 and for all t ∈(0, π) it holds that An(t) n3 −n ≥ An+2(t) (n + 2)3 −(n + 2). (7)
Lemma 4. For all integers n ≥2 and for all t ∈(0, π) it holds that An(t) n3 −n ≥ An+2(t) (n + 2)3 −(n + 2). (7)
Lemma 3
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Lemma 3) we may restrict our attention to t in [0, π]. Suppose first that either the hypothesis (a) or (b) holds, that is, m and n are…
Lemma 3) we may restrict our attention to t in [0, π]. Suppose first that either the hypothesis (a) or (b) holds, that is, m and n are simultaneously odd or even. Note that ϕmn(0) = ϕmn(π) = n3 −n m3 −m for odd m, n and that ϕmn(0) = n3 −n m3 −m < n m = ϕmn(π) for even m, n. Hence, our goal is to show that An(t) Am(t) ≥n3 −n m3 −m
Theorem 1.
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Theorem 1. 4. Appendix: Calculation of qn Here our starting point will be Bombieri’s formula (4.1) in [2]. According to it, if φ is a…
Theorem 1. 4. Appendix: Calculation of qn Here our starting point will be Bombieri’s formula (4.1) in [2]. According to it, if φ is a function in L2[0, 1] then a second variation of the Koebe function is given by q(z) = Q K(z) , where Q(w) = −w2 Z 1 0 φ(u)2 U du −2w3 Z 1
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