Some Inequalities Involving the Derivative of Rational Functions

In this paper


Introduction
Te modulus of complex polynomials on a circle and the locations of zeros of these polynomials have been studied for several years.We start with a result due to Bernstein [1].Let P n be the class of polynomials of degree at most n.If p(z) is a polynomial of degree n, then the famous result, known as Bernstein Equality holds in (1) if and only if p(z) has all zeros at the origin.
Erd € os conjectured in 1944 which Lax [2] proved by improving (1) that for polynomials p(z) of degree n and having no zeros in |z| < 1, we have (3)
Let P m be the class of all polynomials of degree at most m.Now we defne R m,n by R m,n � R m,n a 1 , a 2 , . . ., a n  ≔ p(z) w(z) : p ∈ P m and m ≤ n  . ( Tus, R m,n is the set of all rational functions with poles a 1 , a 2 , . . ., a n at most and with fnite limit at ∞.
From now on, we denote T k � z: |z| � k { }, D − k is the set of all points inside T k , and D + k is the set of all points outside T k .
In 1995, Li et al. [8] extended Bernstein-type inequalities to a rational function by replacing z n by Blaschke product B(z).Tey obtained the following results.
Li et al. [8] also proved the following results.
Theorem 2. If r(z) ∈ R n,n and all the zeros of r(z) lie in for z ∈ T 1 .
Theorem 4. Suppose r(z) ∈ R m,n where r(z) has exactly n poles a 1 , a 2 , . . ., a n and all the zeros of r(z) lie in where m is the number of zeros of r(z).
An extension of Teorem 4 was shown by Wali and Shah [9].Theorem 5. Suppose r(z) ∈ R m,n where r(z) has exactly n poles a 1 , a 2 , . . ., a n and all the zeros of r(z) lie in where m is the number of zeros of r(z).

Main Results
We use the following lemmas for proving our main results.Te frst lemma was shown by Dubinin [10] (see also [11]).
Lemma 7 was proved by Aziz and Zargar [12].Te next lemma is due to Chan and Malik [13] and Li et al. [8].
In this paper, frstly, we obtain an inequality for the modulus of the derivative of rational functions with prescribed poles and restricted zeros.Theorem 10.Suppose r(z) � (p(z)/w(z)) ∈ R m,n where r(z) has exactly n poles a 1 , a 2 , . . ., a n and all the zeros of r(z) lie in T k ∪ D + k , k ≥ 1 except the zeros of order s lying in the origin.Ten, for z ∈ T 1 , where Proof.Let r(z) � (p(z)/w(z)) where p(z) � z s h(z) and h(z Tis implies that Hence,

International Journal of Mathematics and Mathematical Sciences
Since h(z) has all its zeros in From (18), we get for z ∈ T 1 .Hence, for |z| � 1 and using (18), we have for z ∈ T 1 .Combining this with Lemma 9, we get for z ∈ T 1 .Tis completes the proof.□ By taking s � 0 in Teorem 10, we get the next result.
Corollary 11.Suppose r(z) � (p(z)/w(z)) ∈ R m,n where r(z) has exactly n poles a 1 , a 2 , . . ., a n and all the zeros of r(z) By taking k � 1 in Corollary 11, we get the following result.
Corollary 12. Suppose r(z) � (p(z)/w(z)) ∈ R m,n where r(z) has exactly n poles a 1 , a 2 , . . ., a n and all the zeros of r(z) lie in where m is the number of zeros of r(z).
If r(z) ∈ R m,m in Corollary 12, it follows that Corollary 12 reduces to Teorem 2.

International Journal of Mathematics and Mathematical Sciences
Ten, w(z) � (z − α) n and r(z) � (p(z)/(z − α) n ).We get where is the polar derivative of polynomial p(z) with respect to the point α.It generalizes the ordinary derivative in the sense that Now, the next result is obtained for the polar derivative.

Corollary 14. Suppose p(z) � z s h(z) where h(z)
, except the zeros of order s lying in the origin.Ten, for z ∈ T 1 .
Dividing both sides of the inequality (30) by |α| and letting |α| ⟶ ∞, we get the following result of Kumar and Lal [14].

Corollary 15. Suppose p(z) � z s h(z) has all the zeros lying in
Theorem Proof.Let r(z) � (p(z)/w(z)) where p(z) � (z − z 0 ) s h(z) and h(z) �  m− s j�0 c j+s z j is a polynomial of degree m − s having all its zeros in T 1 ∪ D − 1 .By diferentiating with respect to z, we get It is easy to see that Tis implies that Hence, Re zr ′ (z) r(z) Lemma 6 and Lemma 7 yield has the zeros z 0 of order s 0 and the zeros z 1 of order By diferentiating with respect to z, we obtain Te reverse triangle inequality implies that Applying Teorem 16 to r 0 (z), we have for z ∈ T 1 .Tus, the proof is complete.

Conclusions
Tis paper gives an upper bound of a modulus of derivative of rational functions.
where r(z) has exactly n poles a 1 , a 2 , . . ., a n and all the zeros of r(z) lie in T k ∪ D + k , k ≥ 1 except the zeros of order s lying in the origin.Moreover, we give a lower bound of a modulus of derivative of rational functions.
16. Suppose r(z) � (p(z)/w(z)) ∈ R m,n where p(z) � (z − z 0 ) s h(z) has the zeros z 0 of order s with |z 0 | > 1 and h(z) �  m− s j�0 c j+s z j has all its zeros lying in T 1 ∪ D − 1 .Ten, [15]rk 17.From the conditions of Teorem 16, we have|z 0 | > 1. Tus, (1 − |z 0 |/1 + |z 0 |) s < 1 for s > 0.Tat is, our lower bound in Teorem 16 is better than the lower bound in Teorem 3 of Mir et al.[15].From Teorem 16, we obtain the following results in term of polar derivative.Suppose p(z) ∈ P m has all its zeros lying in T 1 ∪ D − 1 .Ten, for any complex number α with |α| ≥ 1, Dividing both sides of the inequality (40) by |α| and letting |α| ⟶ ∞, we get the following result.Suppose p(z) � (z − z 0 ) s h(z) has the zeros z 0 of order s with |z 0 | > 1 and h(z) �  m− s j�0 c j+s z j has all its zeros lying in T 1 ∪ D − 1 .Ten, Corollary 18. Suppose p(z) � (z − z 0 ) s h(z) has the zeros z 0 of order s with |z 0 | > 1 and h(z) �  m− s j�0 c j+s z j has all its zeros lying in T 1 ∪ D − 1 .Ten, for any complex number α with |α| ≥ 1, lower bound of |r 0 ′ (z)| is obtained by Teorem 16.We get a lower bound of |r(z)| as in Corollary 21. 6 International Journal of Mathematics and Mathematical Sciences Next, we can fnd a lower bound of |r i ′ (z)| for 1 ≤ i ≤ v by a similar process by using a lower bound of |r i− 1 ′ (z)| from the previous process and the fact           − n − m + s 0 + s 1 + • • • + s i  + c m         − c s 0 +s 1 +•••+s i           c m         + c s 0 +s 1 +•••+s i           + 2 s 0 + s 1 + • • • + s i  1 + z 0 for 1 ≤ i ≤ v.