【摘要】 讨论了亚纯函数的唯一性问题 ,得到如下结果 :设S ={z|azn-n(n - 1)z2 + 2n(n - 2 )bz -(n - 1) (n - 2 )b2 =0 } ,其中n(>4 )是一个整数 ,a和b是两个非零复数 ,且满足abn - 2 ≠ 2 .如果f与g为非常数亚纯函数 ,且满足E(S ,f) =E(S ,g) ,E({∞ } ,f) =E({∞ } ,g) ,及E({ 0 } ,f) =E({ 0 } ,g) ,则f =g ,或 (f-b) (g -b) =b2 .更多还原

【Abstract】 The uniqueness of meromorphic functions sharing sets is considered,and the following result is obtained:Let S={z|az n-n(n-1)z 2+2n(n-2)bz-(n-1)(n-2)b 2=0},where n(>4),a,bsuch that ab≠0 and ab n-2 ≠2. If f and g are two nonconstant meromorphic functions satisfying WTBXE(S,f)=E(S,g),E({∞},f)=E({∞},g), and E({0},f)=E({0},g)WTBZ, then WTBXf=gWTBZ, or (f-b)(g-b)=b 2.更多还原