OALib Journal期刊
ISSN: 2333-9721
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IM分担一个值的整函数
, PP. 172-175
Keywords: 整函数,分担值,唯一性
Abstract:
讨论了整函数IM分担一个值的唯一性,并得到了如下唯一性定理设f(z)和g(z)是超越整函数,令n,k(≥2)是两个整数且有n>5k+7.如果(fn(z))(k)和(gn(z))(k)IM分担1,则f(z)=c1ecz,g(z)=c2e-cz,其中c1,c2,c是满足(-1)k(c1c2)nc2k=1的常数,或者f(z)≡tg(z),t是满足tn=1的常数.
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