By Min-Teh Cheng

All papers during this quantity are unique (fully refereed) study reviews through members of the particular software on Harmonic research held within the Nankai Institute of arithmetic. the most subject matters comprise: Wavelets, Singular fundamental Operators, Extemal features, H areas, Harmonic research on neighborhood domain names and Lie teams, and so forth. See additionally :G. David "Wavelets and Singular Integrals on Curves and Surfaces", LNM 1465,1991. FROM THE CONTENTS: D.C. Chang: Nankai Lecture in -Neumann Problem.- T.P. Chen, D.Z. Zhang: Oscillary critical with Polynomial Phase.- D.G. Deng, Y.S. Han: On a Generalized Paraproduct outlined by means of Non-Convolution.- Y.S. Han: H Boundedness of Calderon-Zygmund Operators for Product Domains.- Z.X. Liu, S.Z. Lu: functions of H|rmander Multiplier Theorem to Approximation in genuine Hardy Spaces.- R.L. lengthy, F.S. Nie: Weighted Sobolev Inequality and Eigenvalue Estimates of Schr|dinger Operator.- A. McIntosh, Q. Tao: Convolution Singular necessary Operators on Lipschitz Curves.- Z.Y. Wen, L.M.Wu, Y.P. Zhang: Set of Zeros of Harmonic features of 2 Variables.- C.K. Yuan: at the constructions of in the neighborhood Compact teams Admitting internal Invariant capacity.

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21). Vemos que el ZJ- Z2 , angulo arg - - ZJ- ZJ centra I arg -Z2 ll zl-11 ZJ 0 ZJ = arg (z3 - = arg z2 - z 2) - arg (z3 , - z 1) y el angulo ' determma . d os por un nus. arg z 1 estan mo arco de circunferencia entre los puntos z 1 y z 2 ; Por tanto, scgun cl conocido teorema de geometrla elemental obtenemos X z3 - z2 1 z2 = - arg - . 20 ZJ- ZJ 2 ~ Z1 y Demostrar que si z1 + Z2 + Z3 + Z4 = 0 y lt1 l = lz2l = lz3l = lz41, entonces los puntos Z1, Z2, ZJ, Z4 SOn vecttfes de 31. un rectangulb o coinciden dos a dos.

Tomando puestQ que z +a ::j:. z si a ::j:. 0. Regresa ndo a Ia variable z, obtenemos en vez de (x3, y3) cierto punto (x, y) de Ia recta, obtenemos Ia recta y = ax -(a. = YX2l) q ue pasa por cl origen de coordenadas. Zk Por tanto, = Xte Zt i9 z3 = x 3 e , donde Xj ;;<: 0 (j = 1, 2, 3). A partir de las formulas de Viete i9 1- e' n- =x 3 + qx + 7' 2k7r = =0 (1) 20 obtenemos q = - 24, ei . .. ~ = -ei j. Dado que 8 E .. 2':. ]2 ' · 2.. . ·2h = 1, donde t = -z +z -a . Por constgUtente, tk = e'n(k = 0, n- 1).

2':. ]2 ' · 2.. . ·2h = 1, donde t = -z +z -a . Por constgUtente, tk = e'n(k = 0, n- 1). , n 2 n n n k'lr sen2 n k1r) a ( 1 + i ctg -:;; a 2, (k = 1, n - 1) es decir, todas las ra k es se 11>- Solucion. Las coordenadas de los puntos considerados son z 1 = (1,0), Solucion. Para z ::j:. 0 La ecuaci6n dad a es equivalente a Ia ecuac10n t = - 2 k7r . k'lr k'lr sen - +tsen- cos- encuentran en una misma recta paralela al eje imaginario. -+3 = e _,3 , 38 = -1r. n La tercera formula de Viele tenemos r = - a, a E lit A partir de Ia desigualdad (1) para q = - 24, r = - a se obtiene que a2 ~ 211 , es decir, lal ~ 32/2.

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