A Study on Poles and Zeros and Meromorphic Functions with Non-Zero Derivatives and Related Polynomial Differentials

W.K. Hayman, (1958) has proved the following Theorems. Theorem. Suppose that f(z) is meromorphic in and not of the forms , for a complex . Then if , we have Where is defined as in Lemma, below. Theorem. The only functions f (z) In the plane Meromorphic, So f just has a finite number of zeros and poles formed Where and are polynomials. Of these the only functions for which f and f have no zeros are , where n is a positive integer. Theorem. Suppose the f(z) in the plane is meromorphic, that f,f ‗ and f have no zeros and in addition that f(z) has finite order or more generally that Then, We wish to prove the following interesting improvements of the above mentioned theorems. Theorem: Suppose that is meromorphic in and not of the forms or for a complex . (or equivalently is not of the forms or for some complex). Then if, ,we have Where is defined as in Lemma, below. Lemma. If 1 is a positive integer and is meromorphic and is not a polynomial of degree 1 or less, then Where If and With minor modifications otherwise. Lemma : Where f(z) is meromorphic in plus is not polynomial of degree 1 or less, and if in case R finite, then we have Where is defined as in Lemma, with f(z) instead of . Lemma: If f(z) has meromorphic effect in And is not degree 1 or less of a polynomial, then And

We saw in Theorem, that the only functions meromorphic the plane, for which and have no zeros plus Has only finite pole numbers plus Here we have been unable to prove a theorem of this type with no restriction on the poles, but if we assume also that we can weaken the condition on the poles. Theorem: Suppose that is meromorphic in the plane that and have no zeros, and in addition that finite order or more generally that Then Or Proof: We are assuming f(z) is transcendental, since otherwise the follows from Theorem, set and have by Theorem and Lemma for all r except a set of finite measure And

For a sequence . In fact it is a sequence such that and hence then (14) holds as through the series of interval outside a set of finite measure. Now At the poles of the right hand side remains regular and different form zero. Also by hypothesis and have no zeros so that we deduce that Where P (z) is an integral function. Further in view of (14), P(z) must be polynomial. Now since we may set Where g (z) is an integral function and a simple calculation gives Differentiating, From (15), (16) and (17), we get Thus we deduce that is an integral function such that If P(z) is a constant, we deduce just as in Theorem, that reduces to or So we assume that P(z) has degree at least 1 and shall obtain a contradiction. This contradiction arises from the following three Lemmas Lemma: If P(z) is a non-constant polynomial and , then exits constants , such that if the inequality Holds for large R in the annulus outside a set of circles whose radii are at most C2 Rs. We consider first the case when P(z) = z. For given z let z0 be the nearest to z of the equation and set . Then and the definition of z0 we have . Then Now where and so if we have and so Thus if We have Again since We have Thus Hence if we have Provided that for every root z0 of the equation We now apply this result with our polynomial P(z) instead of z and deduce that if Provided that for every z0 or the equation we have To complete the proof of Lemma we need a subsidiary result Lemma: Suppose that is a polynomial of degree , Then there exist positive constants , such that if , t is any complex number and , the inequality Holds for outside asset of k circles, which depend on t, but whose radii are less than Let be the roots of the equation P (z)=t. Then Hence if we must have For at least one value of . Now if C is sufficiently large we have So that if R > max (4,C), (20) holds for R < |z| < 2R unless But for large t the roots of the equation P(z)=t are given approximately by Where. Thus for large t we have if

some . C yields Where C is a suitable constant. Thus if we deduce By (22). Since there is only k roots z, this proves Lemma We can now complete the proof of Lemma Suppose that P(z) is polynomial satisfying the conditions of Lemma and that Then if and is sufficiently large we have So that (19) holds for all complex Zn, except possibly those for which The number of roots Z0 of the equation satisfying (23) is at most for large R. For each such root Lemma 2.1.5 shows that (19) holds outside a set of k circles of radius at most . Thus the sum of the radius of all these circles for Z0 satisfies (23) is at most As required. This completes the proof Lemma We also need following Valiron‘s result Lemma: If g (z) is a transcendental integral function and is a point such that Where N(r) is the central index of g(z). This lemma now yields contradiction as follows: For our integral function , we know that Applying the above lemma, with q = k+1, and q = k, we get And Using the above results, we are led to As , outside a set of finite logarithmic measure. Hence given e > 0, the inequality Holds for , Except the finite logarithmic measure set. For a set E of values of r in the interval [R, 2R], such that For all larger enough R. Thus, We got the contradiction. This proves the theorems.

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PhD Student, MUIT, Lucknow