Ratio Mathematica Volume 38, 2020, pp. 349-365 Uniqueness of an entire function sharing fixed points with its derivatives Md Majibur Rahaman * Imrul Kaish† Abstract The uniqueness problems of an entire functions that share a nonzero finite value have been studied and many results on this topic have been obtained. In this paper we prove a uniqueness theorem for an entire function, which share a linear polynomial, in particular fixed points, with its higher order derivatives. Keywords: Uniqueness; Entire functions; Fixed points; Sharing; Deriva- tives 2010 AMS subject classifications: 30D35. 1 *Department of Mathematics and Statistics, Aliah University, Kolkata, West Bengal 700160, India; e-mail :majiburjrf107@gmail.com †Department of Mathematics and Statistics, Aliah University, Kolkata, West Bengal 700160, India; e-mail:imrulksh3@gmail.com 1Received on April 19th, 2020. Accepted on June 19th, 2020. Published on June 30th, 2020. doi: 10.23755/rm.v38i0.520. ISSN: 1592-7415. eISSN: 2282-8214. ©Md Majibur rahaman. This paper is published under the CC-BY licence agreement. 349 Md Majibur Rahaman and Imrul Kaish 1 Introduction, Definitions and Results Let f be a non-constant meromorphic function in the open complex plane C. A meromorphic function a = a(z) is called a small function of f if T(r,a) = S(r,f), where T(r,f) is the Nevanlinna characteristic function of f and S(r,f) = ◦{T(r,f)}, as r →∞, possibly outside a set of finite linear measure. Let f and g be two non-constant meromorphic functions and a = a(z) be a polynomial. We say that f and g share a CM if f − a and g − a have the same zeros with same multiplicities. On the other hand, we say that f and g share a IM if f − a and g − a have the same zeros ignoring multiplicities. We express the CM sharing and IM sharing respectively by the notations f = a g = a and f = a ⇔ g = a. Let zk(k = 1, 2, . . .) be zeros of f − a and tk be the multiplicity of the zero zk. If zk(k = 1, 2, . . .) are also zeros of g−a and the multiplicity of the zero zk is at least tk then we use the notation f = a → g = a. For standared definitions and notations of the distribution theory we refer the reader to Hayman [1964]. The problem of uniqueness of meromorphic functions sharing values with their derivatives is a special case of the uniqueness theory of meromorphic func- tion. There are some results related to value sharing. In the begining, Jank, Mues and Volkmann Jank et al. [1986] considered the situation that an entire function shares a nonzero value with its derivatives and they prove the following result. Theorem A. Jank et al. [1986]. Let f be a non-constant entire function and a be a non-zero finite value. If f, f(1) and f(2) share a CM, then f ≡ f(1). Following example shows that in Theorem A the second derivative cannot be replaced by any higher order derivatives. Example 1.1. Let k(≥ 3) be an integer and ω(6= 1) is a (k−1)th root of unity. We put f = eωz + ω − 1. Then f, f(1) and f(k) share the value ω CM, but f 6≡ f(1). On the basis of this example, Zhong improved Theorem A by considering higher order derivetives in the following way. Theorem B. Let f be a non-constant entire function and a be a non-zero finite number. Also let n(≥ 1) be a positive integer. If f and f(1) share the value a CM, and if f(n)(z) = f(n+1)(z) = a whenever f(z) = a, then f ≡ f(n). In 2002, Chang and Fang [2002] extendeed Theorem A by considering shared fixed points. 350 Uniqueness of an entire function sharing fixed points with its derivatives Theorem C. Chang and Fang [2002]. Let f be a non-constant entire function. If f, f(1) and f(2) share z CM, then f ≡ f(1). Later in 2003, Wang and Yi [2003] improved Theorem A and generalize The- orem B by considering higher order derivatives in the following way. Theorem D. Wang and Yi [2003]. Let f be a non-constant entire function and a be a non-zero finite constant. Also let m and n be positive integers satisfying m > n. If f and f(1) share the value a CM, and if f(m)(z) = f(n)(z) = a whenever f(z) = a, then f(z) = Aeλz + a− a λ , where A(6= 0) and λ are constants satisfying λn−1 = 1 and λm−1 = 1. In this paper we improve Theorem D by considering the situation when a non- constant entire function f shares a linear plynomial a(z) = αz + β, α( 6= 0) and β are constants, with higher order derivatives. The main result of the paper is the following theorem. Theorem 1.1. Let f be a non-constant entire function and a(z) = αz + β be a polynomial, where α(6= 0) and β are constants. Also let m amd n be two positive integers satisfying m > n > 1. If f(z) = a(z) f(1)(z) = a(z) and f(z) = a(z) → f(m)(z) = f(n)(z) = a(z), then f(z) = Cez or f(z) = Ceλz + a(z) − a(z) λ + α(1 −λ) λ2 , where C and λ are non-zero constants. 351 Md Majibur Rahaman and Imrul Kaish 2 Lemmas In this section we state some necessary lemmas. Lemma 2.1. Ngoan and Ostrovskii [1965]. Let f be an entire function of order at most 1 and k be a positive integer, then m ( r, f(k) f ) = o(log r), as r →∞. The above lemma motivates us to prove the following: Lemma 2.2. Let f be an entire function of finite order and k be a positive integer. Then for any small function a(z) with respect to f(z), m ( r, f(k)(z) −a(k)(z) f(z) −a(z) ) = o(log r), as r →∞. Proof. Let g(z) = f(z) −a(z). Then g(k)(z) = f(k)(z) −a(k)(z). Now by Lemma 2.1 and using above equality, we have m ( r, g(k)(z) g(z) ) = o(log r), as r →∞. This implies m ( r, f(k)(z) −a(k)(z) f(z) −a(z) ) = o(log r), as r →∞. This proves the lemma. Lemma 2.3. Clunie [1962]. Let f be a transcendental meromorphic solution of the equation fnP(f) = Q(f), where P(f) and Q(f) are polynomials in f and its derivatives with meromorphic coefficients aj (say). If the total degree of Q(f) is at most n, then m(r,P (f)) ≤ ∑ j m(r,aj) + S(r,f). 352 Uniqueness of an entire function sharing fixed points with its derivatives Lemma 2.4. Chen and Li [2014]. Let a(z) be an entire function of finite order and Q(z) be a non-constant polynomial. If f is an entire solution of the equation f(k) −eQ(z)f = a(z) such that ρ(f) > ρ(a), then ρ(f) = ∞. We use this Lemma to prove the following one. Lemma 2.5. Let f be a non-constant entire function of finite order and a(z) = αz +β be a polynomial, where α(6= 0) and β are constant. Also let k be a positive integer. If f(z) and f(k)(z) share a(z) CM, then f(k)(z) −a(z) f(z) −a(z) ≡ c, (2.1) for some nonzero constant c. Proof. Since f has finite order and since f(z) and f(k)(z) share a(z) CM, it fol- lows from the Hadamard factorization theorem that f(k)(z) −a(z) f(z) −a(z) ≡ eQ(z), (2.2) where Q(z) is a polynomial. Suppose that F(z) = f(z) −a(z). Then F(k)(z) = f(k)(z). From (2.2) and above equality, we have F(k)(z) −eQ(z)F(z) = a(z). If Q(z) is non-constant, then from above equality and by Lemma 2.4, we get F has infinite order. Since f has finite order, this is impossible. Hence Q(z) is a constant. Therefore from (2.2), we obtain (2.1) for a non-zero constant c. This proves the lemma. Lemma 2.6. Let f be a transcendental entire function of finite order and a(z) = αz + β be a polynomial, where α(6= 0) and β are constants. Also let m be a positive integer. If (i) m ( r, 1 f(z)−a(z) ) = S(r,f), (ii) f(z) = a(z) f(1)(z) = a(z) and (iii) f(z) = a(z) → f(m)(z) = a(z), then f(z) = Cez, where C is a non-zero constant. 353 Md Majibur Rahaman and Imrul Kaish Proof. Let h(z) = f(1)(z) −a(z) f(z) −a(z) . (2.3) Since f(z) and f(1)(z) share a(z) CM, we see that h(z) is an entire function. Now by Lemma 2.1, Lemma 2.2 and from the hypothesis of Lemma 2.6, we deduce that T(r,h(z)) = m(r,h(z)) = m ( r, f(1)(z) −a(z) f(z) −a(z) ) ≤ m ( r, f(1)(z) −a(1)(z) f(z) −a(z) ) + m ( r, a(1)(z) −a(z) f(z) −a(z) ) + log 2 = S(r,f). (2.4) We rewrite (2.3), as f(1)(z) = h(z)f(z) + a(z)(1 −h(z)) = ξ1(z)f(z) + η1(z), (2.5) where ξ1(z) and η1(z) are defined by ξ1(z) = h(z), η1(z) = a(z)(1 −h(z)). By (2.5), we have f(2)(z) = ξ1(z)f (1)(z) + ξ (1) 1 (z)f(z) + η (1) 1 (z) = ξ1(z)[ξ1(z)f(z) + η1(z)] + ξ (1) 1 (z)f(z) + η (1) 1 (z) = [ξ (1) 1 (z) + ξ1(z)ξ1(z)]f(z) + η (1) 1 (z) + η1(z)ξ1(z) = ξ2(z)f(z) + η2(z), where ξ2(z) = ξ (1) 1 (z) + ξ1(z)ξ1(z) and η2(z) = η (1) 1 (z) + η1(z)ξ1(z). Now from above equality and using (2.5), we get f(3)(z) = ξ2(z)f (1)(z) + ξ (1) 2 (z)f(z) + η (1) 2 (z) = [ξ (1) 2 (z) + ξ1(z)ξ2(z)]f(z) + η (1) 2 (z) + η1(z)ξ2(z) = ξ3(z)f(z) + η3(z), 354 Uniqueness of an entire function sharing fixed points with its derivatives where ξ3(z) = ξ (1) 2 (z) + ξ1(z)ξ2(z) and η3(z) = η (1) 2 (z) + η1(z)ξ2(z). Similarly, f(k)(z) = ξk(z)f(z) + ηk(z), (2.6) where ξk+1(z) = ξ (1) k (z) + ξ1(z)ξk(z) (2.7) and ηk+1(z) = η (1) k (z) + η1(z)ξk(z). (2.8) Puting k = 1 in (2.7), we have ξ2(z) = ξ (1) 1 (z) + ξ1(z)ξ1(z) = h2(z) + h(1)(z). Again puting k = 2 in (2.7), we get ξ3(z) = ξ (1) 2 (z) + ξ1(z)ξ2(z) = [ h2(z) + h(1)(z) ](1) + h(z)[h2(z) + h(1)(z)] = h3(z) + h(2)(z) + 3h(z)h(1)(z). Similarly, ξ4(z) = h 4(z) + h(3)(z) + 4h(z)h(2)(z) + 3 [ 2h2(z) + h(1)(z) ] h(1)(z). Hence using mathematical induction, one can easily check ξk(z) = h k(z) + Pk−1(z,h(z)), (2.9) where Pk−1(z,h(z)) is a polynomial such that total degree degPk−1(z,h(z)) ≤ k−1 in h(z) and its derivatives, and all coefficients in Pk−1(z,h(z)) are constants. Now putting k = 1 in (2.8), we have η2(z) = η (1) 1 (z) + η1(z)ξ1(z) = [a(z)(1 −h(z))](1) + a(z)(1 −h(z))h(z) = −a(z)h2(z) −a(z)h(1)(z) + (a(z) −α)h(z) + α. 355 Md Majibur Rahaman and Imrul Kaish Again putting k = 2 in (2.8), we get η3(z) = η (1) 2 (z) + η1(z)ξ2(z) = [ −a(z)h2(z) −a(z)h(1)(z) + (a(z) −α)h(z) + α ](1) +a(z)(1 −h(z))(h2(z) + h(1)(z)) = −a(z)h3(z) −a(z)h(2)(z) + [2a(z) − 3a(z)h(z) − 2α] h(1)(z) +(a(z) −α)h2(z) + αh(z). Similarly, = −a(z)h4(z) −a(z)h(3)(z) + [3a(z) − 4a(z)h(z) − 3α] h(2)(z) + [ 5a(z)h(z) − 5αh(z) − 6a(z)h2(z) − 3a(z)h(1)(z) + 3α ] h(1)(z) +(a(z) −α)h3(z) + αh2(z). Like the previous one, it can be easily verified that ηk(z) = −a(z)hk(z) + Qk−1(z,h(z)), (2.10) where Qk−1(z,h(z)) is a polynomial such that total degree degQk−1(z,h(z)) ≤ k − 1 in h(z) and its derivatives, and all coefficients in Qk−1(z,h(z)) are either constants or polynomial a(z). From (2.4) and (2.9), for k = 1, 2, · · · , we have T(r,ξk(z)) = T(r,h k(z) + Pk−1(z,h(z))) ≤ T(r,hk(z)) + T(r,Pk−1(z,h(z))) + log 2 = S(r,f). Similarly, T(r,ηk(z)) = S(r,f). From hypothesis of Lemma 2.6, we have N ( r, 1 f(z) −a(z) ) = T(r,f(z)) −m ( r, 1 f(z) −a(z) ) + O(1) = T(r,f(z)) + S(r,f), (2.11) which implies that f(z) −a(z) must have zeros. Let zj be a zero of f(z) −a(z) with multiplicity δ(j). Since f(z) = a(z) → f(m)(z) = a(z), we see that zj is also a zero of f(m)(z) −a(z) with multiplicity at least δ(j). Hence f(zj) = a(zj) and f(m)(zj) = a(zj). 356 Uniqueness of an entire function sharing fixed points with its derivatives It follows from (2.6) that, for k = m, f(m)(z) = ξm(z)f(z) + ηm(z) (2.12) and then a(zj) = a(zj)ξm(zj) + ηm(zj). Now we shall prove that, a(z) ≡ a(z)ξm(z) + ηm(z). (2.13) Otherwise, a(z)ξm(z) + ηm(z) −a(z) 6≡ 0. From (2.12), we have a(z)ξm(z) + ηm(z) −a(z) = (f(m)(z) −a(z)) − ξm(z)(f(z) −a(z)). By the reasoning as mentioned above, we deduce that zj is a zero of (f(m)(z)− a(z)) − ξm(z)(f(z) − a(z)), that is, a zero of a(z)ξm(z) + ηm(z) − a(z) with multiplicity at least δ(j). It follows from this and the fact that ξm(z) and ηm(z) are small functions of f(z), N ( r, 1 f(z) −a(z) ) ≤ N ( r, 1 a(z)ξm(z) + ηm(z) −a(z) ) ≤ T ( r, 1 a(z)ξm(z) + ηm(z) −a(z) ) = S(r,f), which contradicts (2.11). Thus a(z) ≡ a(z)ξm(z) + ηm(z), which is (2.13). Now by induction we prove that ηk+1(z) + a(z)ξk+1(z) = (a(z) −α)hk(z) + Rk−1(z,h(z)), (2.14) where Rk−1(z,h(z)) is a polynomial such that degRk−1(z,h(z)) ≤ k− 1 in h(z) and its derivatives, and all the coefficients in Rk−1(z,h(z)) are constants or poly- nomial a(z). 357 Md Majibur Rahaman and Imrul Kaish Firstly, from (2.7), (2.8) and for k = 1, we have η2(z) + a(z)ξ2(z) = η (1) 1 (z) + η1(z)ξ1(z) + a(z) [ ξ (1) 1 (z) + ξ1(z)ξ1(z) ] = [a(z)(1 −h(z))](1) + a(z)(1 −h(z))h(z) + a(z)h(1)(z) +a(z)h2(z) = a(z)(−h(1)(z)) + α(1 −h(z)) + a(z)h(z) −a(z)h2(z) +a(z)h(1)(z) + a(z)h2(z) = (a(z) −α)h(z) + α. Secondly, we suppose that the following equation holds ηk(z) + a(z)ξk(z) = (a(z) −α)hk−1(z) + Rk−2(z,h(z)). Now, by (2.7)–(2.10), we deduce that ηk+1(z) + a(z)ξk+1(z) = η (1) k (z) + η1(z)ξk(z) + a(z)(ξ (1) k (z) + ξ1(z)ξk(z)) = [ −a(z)hk(z) + Qk−1(z,h(z)) ](1) + a(z)(1 −h(z))ξk(z) +a(z) [ hk(z) + Pk−1(z,h(z)) ](1) + a(z)h(z)ξk(z) = −ka(z)hk−1(z) −αhk(z) + [Qk−1(z,h(z))] (1) + a(z)ξk(z) −a(z)h(z)ξk(z) + ka(z)hk−1(z) + a(z) [Pk−1(z,h(z))] (1) +a(z)h(z)ξk(z) = a(z) [ hk(z) + (Pk−1(z,h(z))) ] −αhk(z) + [Qk−1(z,h(z))] (1) +a(z) [Pk−1(z,h(z))] (1) = (a(z) −α)hk(z) + Rk−1(z,h(z)), which proves (2.14). From (2.13) and (2.14), we obtain (a(z) −α)hm−1(z) + Rk−2(z,h(z)) ≡ a(z). (2.15) Clearly, Rk−2(z,h(z)) 6≡ a(z). Othewise, from (2.3), (2.15) and the hypothesis of Lemma 2.6, we have a contradiction. Hence by Lemma 2.3 and from (2.15), we can deduce that h(z) must be constant. From (2.7) and ξ1(z) = h(z), we have ξ2(z) = h 2(z), ξ3(z) = h 3(z), ξ4(z) = h 4(z). Similarly, ξk(z) = h k(z), for k = 1, 2, . . . . (2.16) 358 Uniqueness of an entire function sharing fixed points with its derivatives Now, from (2.8) and η1(z) = a(z)(1 −h(z)), we get η2(z) = (1 −h(z))(α + a(z)h(z)), η3(z) = (1 −h(z))(α + a(z)h(z))h(z), η4(z) = (1 −h(z))(α + a(z)h(z))h2(z). Similarly, ηk(z) = (1 −h(z))(α + a(z)h(z))hk−2(z), for k = 2, 3, . . . . (2.17) From (2.13), (2.16) and (2.17), we have a(z) ≡ a(z)hm(z) + (1 −h(z))(α + a(z)h(z))hm−2(z) ≡ hm−2(z) [ a(z)h2(z) + α(1 −h(z)) + a(z)h(z) −a(z)h2(z) ] ≡ hm−2(z) [a(z)h(z) + α(1 −h(z))] , which implies that h(z) = 1. Hence from (2.3) and h(z) = 1, we can obtain f(1)(z) = f(z). This implies f(z) = Cez. where C(6= 0) ia a constant. This proof the Lemma 2.6. 3 Proof of the theorem 1.1 First we verify that f(z) cannot be a polynomial. Let f(z) be a polynomial of degree 1. Suppose that f(z) = A1z + B1, where A1(6= 0) and B1 are constants. Then f(1)(z) = A1, f(m)(z) ≡ 0 ≡ f(n)(z). Now β−B1A1−α is the only zero of f(z) − a(z), A1−β α is the only zero of f(1)(z) − a(z) and −β α is the only zero of f(m)(z)−a(z). Since f(z) and f(1)(z) share polynomial a(z) CM and the zeros of f(z) −a(z) are the zeros of f(m)(z) −a(z), we have A1−β α = −β α and so A1 = 0, a contradiction. Now let f(z) be a polynomial of degree greater than 1. Suppose that deg(f(z)) = p. Then deg(f(z)−a(z)) = p and deg(f(1)(z)−a(z)) = p−1, it contradicts the fact that f(z) and f(1)(z) share polynomial a(z) CM. Hence f(z) is a transcendental entire function. Thus T(r,a(z)) = S(r,f). 359 Md Majibur Rahaman and Imrul Kaish To prove the theorem let us consider two functions defined as follows. Φ(z) = (a(z) −a(1)(z))f(m)(z) −a(z)(f(1)(z) −a(1)(z)) f(z) −a(z) (3.1) and Ψ(z) = (a(z) −a(1)(z))f(n)(z) −a(z)(f(1)(z) −a(1)(z)) f(z) −a(z) . (3.2) Then Φ(z) 6≡ Ψ(z). We know from the hypothesis of Theorem 1.1 that Φ(z) and Ψ(z) are entire functions. Then, by Lemma 2.1 and Lemma 2.2, we have T(r, Φ(z)) = m(r, Φ(z)) = m ( r, (a(z) −a(1)(z))f(m)(z) −a(z)(f(1)(z) −a(1)(z)) f(z) −a(z) ) ≤ m ( r, (a(z) −a(1)(z)) f(m)(z) f(z) −a(z) ) + m ( r,a(z) (f(1)(z) −a(1)(z)) f(z) −a(z) ) + log 2 = S(r,f). Similarly, T(r, Ψ(z)) = S(r,f). We shall the following three cases. Case 1. First we suppose that Φ(z) 6≡ 0. Then by (3.1), we have f(z) = a(z) + 1 Φ(z) {(a(z) −a(1)(z))f(m)(z) −a(z)(f(1)(z) −a(1)(z))}. (3.3) From (3.1) and (3.2), we get f(1)(z) = (a(z) −a(1)(z)) a(z)(Φ(z) − Ψ(z)) (Φ(z)f(n)(z) − Ψ(z)f(m)(z)) + a(1)(z). Therefore f(1)(z) −a(z) = (a(z) −a(1)(z)) a(z)(Φ(z) − Ψ(z)) (Φ(z)f(n)(z) − Ψ(z)f(m)(z)) +a(1)(z) −a(z). (3.4) 360 Uniqueness of an entire function sharing fixed points with its derivatives First we suppose that m > n > 2. Then from (3.4), we get 1 f(1)(z) −a(z) = 1 a(z)(Φ(z) − Ψ(z)) (Φ(z)f(n)(z) − Ψ(z)f(m)(z)) f(1)(z) −a(z) + 1 a(1)(z) −a(z) . (3.5) Using Lemma 2.1 and from (3.5), we have m ( r, 1 f(1)(z) −a(z) ) = S(r,f). (3.6) Next we suppose m > n = 2. Then from (3.4), we get (f(1)(z) −a(z))(Φ(z) − Ψ(z))a(z) = γ(z) + (a(z) −a(1)(z))Φ(z)(f(2)(z) −a(1)(z)) −(a(z) −a(1)(z))Ψ(z)f(m)(z), (3.7) where γ(z) = (a(z) −a(1)(z))(Φ(z)a(1)(z) − (Φ(z) − Ψ(z))a(z)). Clearly γ(z) 6≡ 0. If γ(z) ≡ 0, then Φ(z) ≡ a(z) a(z) −a(1)(z) Ψ(z), which is a contradiction because Φ(z) and Ψ(z) are entire functions and Ψ(z) 6= 0 when a(z) −a(1)(z) = 0. Now from (3.7) we get 1 f(1)(z) −a(z) = (Φ(z) − Ψ(z))a(z) γ(z) − a(z) −a(1)(z) γ(z) Φ(z) f(2)(z) −a(1)(z) f(1)(z) −a(z) + a(z) −a(1)(z) γ(z) Ψ(z) f(m)(z) f(1)(z) −a(z) . (3.8) Again using Lemma 2.1 and from (3.8), we get m ( r, 1 f(1)(z) −a(z) ) = S(r,f), which is (3.6). 361 Md Majibur Rahaman and Imrul Kaish Since Φ(z) 6≡ 0, it follows from (3.3) and Lemma 2.1 that T(r,f(z)) = m(r,f(z)) = m ( r,a(z) + 1 Φ(z) {(a(z) −a(1)(z))f(m)(z) −a(z)(f(1)(z) −a(1)(z))} ) = m ( r,a(z) + (a(z) −α)f(m)(z) −a(z)f(1)(z) + a(z)α Φ(z) ) ≤ m(r,a(z)) + m ( r, (a(z) −α)f(m)(z) −a(z)f(1)(z) Φ(z) ) + m ( r, αa(z) Φ(z) ) + log 3 = m  r,a(z)f(1)(z) (a(z)−α)a(z) f(m)(z)f(1)(z) − 1 Φ(z)   + S(r,f) ≤ m  r, (a(z)−α)a(z) f(m)(z)f(1)(z) − 1 Φ(z)   + m(r,f(1)(z)) + S(r,f) ≤ m ( r, f(m)(z) f(1)(z) − 1 ) + m(r,f(1)(z)) + S(r,f) = T(r,f(1)(z)) + S(r,f). (3.9) Applying Lemma 2.1, We can easily see that T(r,f(1)(z)) = m(r,f(1)(z)) = m ( r, f(1)(z) f(z) ·f(z) ) ≤ m ( r, f(1)(z) f(z) ) + m(r,f(z)) = m(r,f(z)) + S(r,f) ≤ T(r,f(z)) + S(r,f). (3.10) Combining (3.9) and (3.10), we have T(r,f(1)(z)) = T(r,f(z)) + S(r,f). (3.11) Since f(z) and f(1)(z) share a(z) CM, by using (3.6) and (3.11) together with 362 Uniqueness of an entire function sharing fixed points with its derivatives the First Fundamental Theorem, we obtain m ( r, 1 f(z) −a(z) ) = T(r,f(z)) −N ( r, 1 f(z) −a(z) ) + O(1) = T(r,f(1)(z)) −N ( r, 1 f(1)(z) −a(z) ) + S(r,f) = m ( r, 1 f(1)(z) −a(z) ) + N ( r, 1 f(1)(z) −a(z) ) −N ( r, 1 f(1)(z) −a(z) ) + S(r,f) = m ( r, 1 f(1)(z) −a(z) ) + S(r,f) = S(r,f). Hence by Lemma 2.6, we have f(z) = Cez, where C(6= 0) is a constant. Case 2. Now we suppose that Ψ(z) 6≡ 0. Then following the similar argu- ments of Case-1 and using Lemma 2.6, we have f(z) = Cez. where C(6= 0) is a constant. Case 3. Finally we suppose that Φ(z) ≡ 0 and Ψ(z) ≡ 0. Then from (3.1) and (3.2), we get (a(z) −a(1)(z))f(m)(z) −a(z)(f(1)(z) −a(1)(z)) ≡ 0 (3.12) and (a(z) −a(1)(z))f(n)(z) −a(z)(f(1)(z) −a(1)(z)) ≡ 0. (3.13) Now subtracting (3.13) from (3.12), we have (a(z) −a(1)(z))(f(m)(z) −f(n)(z)) ≡ 0. Since a(z) 6≡ a(1)(z), we get f(m)(z) ≡ f(n)(z). 363 Md Majibur Rahaman and Imrul Kaish Solving this we have f(z) = p0 + p1e t1z + p2e t2z + · · · + pm−netm−nz, where t1, t2, · · · , tm−n are distinct (m − n)th roots of unity and p0, p1, p2, · · · pm−n are constants. Since f(z) and f(1)(z) share a(z) CM, applying Lemma 2.5, we get f(1)(z) −a(z) f(z) −a(z) = λ, for some nonzero constant λ. Solving above equality, we obtain f(z) = Ceλz + a(z) − a(z) λ + α(1 −λ) λ2 , where C(6= 0) is a constant. This completes the proof of Theorem 1.1. 4 Conclusions After the above discussion we arrive at the conclusion that if an entire function and its first derivative share a linear polynomial with counting multiplicity and it partially shares the linear polynomial with its two higher order derivatives then the funtion is either one of the following two forms. 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