### Journal Menu >> Applied Mathematics, 2011, 2, 230-235 doi:10.4236/am.2011.22025 Published Online February 2011 (http://www.SciRP.org/journal/am) Copyright © 2011 SciRes. AM Uniqueness of Meromorphic Functions Concerning Differential Monomials* Hui Huang, Bin Huang College of Mathematics, Changsha University of Science and Technology, Changsha, China E-mail: huang.liuyuan@163 .com Received October 21, 2010; revised December 6, 2010; accepted December 11, 2010 Abstract Considering the uniqueness of meromorphic functions concerning differential monomials ,we obtain that, if two non-constant meromorphic functions fz and gz satisfy  1, 1,nnkkEffE gg, where k and n are tow positive integers satisfying 3k and 11n, then either  12,cz czzzfce gce where 1c, 2c, c are three constants, satisfying 1212 1ncc c or ftg for a constant t such that 11nt. Keywords: Meromorphic Function, Sharing Value, Uniqueness 1. Introduction and Main Results In this paper we use the standard notations and terms in the value distribution theory . Let fz be a nonconstant merom orop hi c funct i on on the complex plane C. Define  ,0Eafzf za, where a zero point with multiplicity m is counted m times in the set. If these zero points are only counted once, then we deno te the set by ,Eaf . Let k a pos-itive integer. Define  (),0,,1,.0ikEafzfz aiikstfz , where a zero point whit multiplicity m is counted m times in the set. Let fz and gz be two nonconstant meromor- phic functions. If ,,Eaf Eag, then we say that fz and gzshare the value CM; if ,,Eaf Eag, then we say that fz and gz share the value IM. Additional, we denote by ,kNrf the counting function for poles of fz with multiplicity k, and by kN the corresponding one for which multiplicity is not counted. Let ,kNrf be the counting function for poles of fz with multiplicityk, and by ,kNrf the corresponding one for which multiplicity is not counted. Set  (2,, ,kkN rfNrfNNrf, Similary, we have the notation: 1,kNrf, 1,kNrf, 1,,kNrf 1,,kNrf 1,kNrf. If  1, 1,Ef Eg, we de-note by 11 1,1Nrf the counting function for com- mon simple 1-points of both fz and gz where multiplicity is no t counted. In 1998, Wang and Fang  (cf. ) proved the fol-lowing therem. Theorem A Let fz be a transcendental mero-morphic function, and n, k be tow positive integers with 1nk. Then 1knf has infinitely many zeros. It is interesting to establish the unicity theorem cor-responding to the above result. In 2002, Fang  ob-tained the following result. Theorem B Let ,fg be tow nonconstant entire function, and n, k be tow positive integers with 24nk. If knf and kng share 1 CM, then either 1czzfce, 2czzgce where 1c,2c,c are three constants, satisfying  21211knkcc c, or ftg for a constant t such that 1nt. Recently, Bhoosnurmath and Dyavanal  extended Theorem B to the meromorphic case, as follows. Theorem C Let ,fg be tow nonconstant meromor-*This work is supported by the National Natural Science Foundation of China (Grant No. 11071064) and Hunan Provincial Department of Edu-cation, P.R. of China (No. 05C268). H. HUANG ET AL. Copyright © 2011 SciRes. AM 231phic function, and n, k be tow positive integers with 38nk. If knf and kng share 1 CM, then either  12,cz czzzfcegce where 1c,2c,care three constants, satisfying 21211knkcc c, or ftg for a constant t such that 1nt. Let 1k, 111nfnF and 111ngnG in Theorem C. Then 1nnfFF  and 1nnGg GG . We see that the following result, which is proved by Yang and Hua , is a direct consequence of Theorem C. Theorem D Let fz and gz be two noncons-tant meromorphic functions, and 11n an integer. If nffand nggshare 1 CM, then either 1czzfce, 2czzgce where 1c,2c,c are three constants, satis-fying 1212 1ncc c or ftg for a constant t such that 11nt. In this paper, we will extend the above result as fol-lows. Theorem 1 Let fz and gz be two noncons-tant meromorphic functions, 3k, 11n be tow positive integers. If 1, 1,nnkkEffEgg, then ei- ther  12,cz czzzfcegce where 1c,2c,c are three constants, satisfying 1212 1ncc c or ftg for a constant t such that 11nt. Theorem 2 Let fz and gz be two noncons-tant meromorphic functions, 13n be a positive in- teger. If 221, 1,nnEffEgg, then the conclusion of Theorem 1 holds. Theorem 3 Let fz and gz be two noncons-tant meromorphic functions, 19n be a positive in-teger. If 111, 1,nnEff Egg, then the conclusion of Theorem 1 holds. 2. Some Lemmas For the proof of our results, we need the following lem-mas. Lemma 1 . Let f be a nonconstant meromorphic function,and let 01,,,naa a be finite complex numbers such that 0na. Then 11110,,,nnnnTrafa fafanT r fSr f Lemma 2 . Let fz and gz be two non-constant meromorphic functions, 6n be a positive integer, if 1nnffgg then either 1czzfce, 2czzgce where 1c,2c,c are three constants, satis-fying 1212 1ncc c. Lemma 3 . Let f be a nonconstant meromorphic function, k a positive integer, then 11,,,,kNrNrkNrf Srfff . Lemma 4 . Let f and g be two nonconstant meromorphic functions,and let k be a positive integer. If 1, 1,kkEfEg, then one of the following cases must occur:  2222 2211 111111,, ,,, ,,,11111,(,),,,111kkTrfTrgNrfNrNrg NrNrNrfgfgNrNrNrSrf Srgfff            (1)   11,where 0, are tow constants.bgabfabbga b (2) Lemma 5. Let f and g be two nonconstant me-romorphic functions, 6n be a positive integer, set nFff, nGgg, if 11bGabFbGa b (2.1) where 0ab are two constants, then either 1czzfce, 2czzgce where 1c,2c,c are three constants, satis-fying 1212 1ncc c or ftg for a constant t such that 11nt. Proof. By Lemma 1, we get   '',,, ,,2,,2,,nnTrFTrffTrfTrfnTrfTrfSrfnTrfSrf  (2.2) H. HUANG ET AL. Copyright © 2011 SciRes. AM 232   ,,, ,,,1,,,,,1,,,,,1,,, ,,nnnnnnnTrfTrf SrfNrf mrf SrfNrf fNrfmrffmrSrffTrffTrf Nrf NrSrffTrFTrfNrfNrSrff    (2.3) So,  1,1,,, ,TrFnTrf Nrf NrSrff  (2.4) Thus, by (2.2 ) and (2.3) , we ge t ,,SrF Srf. Similarly, we get  ,1,,1,,TrGnTrg NrgNr Srgg  (2.5) ,,SrG Srg It is clear that the inequality ,,Trf Trg or  ,,Trg Trf holds for a set of infinite measure of r. Without loss of generality, we may suppose that ,,TrfTrg, holds for rI, where I is a set with infinite measure. Next we consider five cases. Case 1. ,0,1abb, If 10ab , then by the 2.1 we known: 11,, 11NrNr abFGb   By the Nevalinna second fundamental theorem and lemma 3, we have      '11,,, ,,1111,, ,,11 11,, ,,, ,11 1,, ,,2,,TrGNrG NrNrSrGabGGbNrG NrNrSrgGFNrg NrNrNrNrSrgggf fNrg NrNrNrfNrSrggg f               1,,,3,,14(,),,,TrgNrgNrTrfSrggTrgNrgN rSrgg  By 6n and (2.5), we get ,,Trg Srg, for rI, a contradiction. If 10ab, by (2.1 ) w e can obtain: 11bGFbG We see that: 1,,1NrF NrGb Combining the Nevalinna second fundamental theo-rem and lemma 3, we have H. HUANG ET AL. Copyright © 2011 SciRes. AM 233        11,,, ,,11,,,,111,, ,,,1,,,,,12(,),,,TrGNrGNrNrSrGGGbNrG NrNrF SrgGNrg NrNrNrSrggg fTrg NrgNrTrfSrggTrgNrgN rSrgg        By 6n and (2.5), we get  ,,Trg Srg, rI, a contradiction. Case 2. ,1abb, So 1aFaG We can get  1,,1NrF NrGa, similarly as Case 1, it is impossible. Case 3. ,0abb, So 1GaFa. If 10a , then FG, so nnffgg. It follows that: 111111nnnn fgffggFC GCnn , where C is a constant. We state that C is zero. If not, one we can get that from the Nevalinna second fundamental theorem and lemma 1.    11111111, ,,11,, ,,111,, ,,11,, ,,3, ,nTrgT rGSrgNrG NrNrSrgGGNrG NrNrSrgGFNrg NrNrSrggfTrg Srg       Because 6n, we can get  ,,Trg Srg, for rI, which is impossible. So C is zore. Then 11FG, it gives that 11nnfg, so ,ftg where t is constant satisfying 11nt. Case 4. ,0,1abb, from (2.1) we can get 11bGFbG 1,(,)NrF NrG, similarly as Case 1, it is impossible. Since 0a, now we consider the following case. Case 6. 1ab It yields 1FG, that is: 1nnffgg. By the Lemma 2, we can get  12,cz czzzfce gce where 1c,2c,c are three constants, satisfying 1212 1ncc c . Now the proof of Lemma 5 is completed. 3. Proof of Theorems Proof of theorem 1: Noticing that 3k, we have  11(1 (111 1,, ,11 111,,111111,,212111,,122kkNrNrNrFG FNr NrFGNr NrFGTrFTrGO By lemma 4, we can get     2222222211,,2,,,,,,112,,,,, ,T rFT rGNrNrFNrNrGSrFS rGFGNrNrF NrNrGSrfSrgFG        (3.1) H. HUANG ET AL. Copyright © 2011 SciRes. AM 234 Because: 222211 11,,, ,2,,2,nnNrN rFNrN rffNrNrNrfFffff         (3.2) and  22111,,2,,2,NrNrGNrNrNrgGgg       (3.3) By (3.1)-(3.3) and lemma 3, we can get:      1111,,22,2,,2,2, ,,,11114,4, 2,, 4,4,2,,115, 5,,TrF TrGNrNrfNrNrNrgNrSrfSrgffggNrNrfNrSrfNrNrgNrSrgffggNrNrf NrSff                11,5,5,,,119,,,, 9,,,,rfNrNrgNrSrgggT rfNrfNrSrfT rgN rgNrSrgfg        (3.4) By 9nand (2.4), (2.5) we obtain ,,,,T rfT rgSrfSrg, which is impossi-ble. Therefor e, by lemma 4  11bgabfbga b, where 0a, b are tow constants, it follows by lemma 5 then either  12,cz czzzfce gce where 1c,2c,c are three constants, satisfying 1212 1ncc c or ftg for a constant t such that 11nt. The proof of Theorem 1 is complete. Proof of theorem 2: We can see clearly:  (3 (31111 11111,,,,,11 1212111111 1,,,,,,21212 2NrNrNrNrN rFG FFGNrNrTrFTrGSrfSrgFG    By lemma 4, we can get:  2222(3 (311,,2,, ,,11,,,,11TrFTrGNrN rFNrN rGFGNr NrSrfSrgFG     (3.5) Considering   '(3 ''1111 11,,,,,,,1222 211 1,,,,2,,2FFN rNrNrSrfNrNrF SrfFFFFNrNrNrfSrfTrf Srfff    (3.6) Similarly, we can get (3 1,2,,1Nr TrgSrgG (3.7) By from (3.4)-(3.7), we can get   11,,11,,,,11 ,,,,TrFTrGTrfNrfNrSrfTrg NrgNrSrgfg    H. HUANG ET AL. Copyright © 2011 SciRes. AM 235 Since 13n and (2.4), (2.5), we can get ,,,,T rfT rgSrfSrg impossible The proof of Theorem 2 is complete. Proof of theorem 3: Since:  1111 111,, ,,11 121111 1,,,,,212 2NrNrN rNrFG FFNrTrFTrGSrfSrgG   We can see clearly from lemma 4 that:    (2 (22222(2 (2222211 11,,2,, ,,,,11,,11 112,, ,,,,11,,T rFTrGNrNrFNrNrGNrNrFG FGSrF SrGNr NrFNr NrGNrNrFG FGSrf Srg          (3.8) Considering    (2 11,,,,,,,111,,,,4,,FFN rNrNrSrfNrFNrSrfFFF FNrNrNrf SrfTrf Srfff        (3.9) Similarly, we can get (2 1,4,,1Nr TrgSrgG (3.10) By from (3.8)-(3.10), we can get    11,,17.,,,17 .,,,TrFTrGTrfNrfNrSrfTrgNrgNrSrgfg    Since 19n and (2.4), (2.5), we can get ,,,,T rfT rgSrfSrg , impossible The proof of Theorem 3 is complete. 4. References  L. Yang, “Value Distribution Theory,” Springer-Verlag, Berlin, 1993.  Y. F. Wang and M. L. 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