AE03401720180

International Journal of Computational Engineering Research||Vol, 03||Issue, 4||

On The Zeros of Polynomials and Analytic Functions M. H. Gulzar Department of Mathematics University of Kashmir, Srinagar 190006

Abstract In this paper we obtain some results on the zeros of polynomials and related analytic functions, which generalize and improve upon the earlier well-known results.

Mathematics Subject Classification: 30C10, 30C15 Key-words and phrases: Polynomial, Analytic Function , Zero.

I.

INTRODUCTION AND STATEMENT OF RESULTS

Regarding the zeros of a polynomial , Jain [2] proved the following results: Theorem A: Let P ( z )  a 0  a 1 z  ......  a p  1 z n such that a

 a

p 1

p

p 1

 apz

p

 ......  a n z

n

be a polynomial of degree

for some p  1, 2 ,......, n  ,

n

M  M





p

a

 a

j

j 1

 an ,

(1  p  n ), M

n

 an

j  p 1

p 1

m  m

p





a

 a

j

j 1

( 2  p  n ), m 1  0 .

,

j 1

Then P(z) has at least p zeros in p (a

z 

 a

p

)

p 1

,

p 1

M

provided p (a

p

 a

p 1

)

p 1

M

1

and a0

p  a p  a p 1   m M  p 1

  p   a p  a p 1      M p 1     p

  .  

Theorem B: Let P ( z )  a 0  a 1 z  ......  a p  1 z such that a

p 1

 a

 apz

p

 ......  a n z

n

be a polynomial of degree n

for some p  1, 2 ,......, n  ,

p

   

p 1



, j  0 ,1,......., n , 2 for some real  and  and arg a

j

a n  a n  1  ......  a 1  a 0 .

Then P(z) has at least p zeros in z 

p a L

p

 a

p 1

p 1

,

where n

L  L

p

 an  ( an  a

p

) cos  



(a

j

 a

j 1

) sin 

j  p 1

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Page 172

On The Zeros Of Polynomials... p 1

l  lp  (a



 a 0 ) cos  

p 1

(a

j

 a

( 2  p  n  1 ), l 1  0 ,

) sin 

j 1

j 1

provided p a

a0  l

p

 a

p  1

L

p

 p      L 

p 1

 a p  a p 1   p  1 

   

p 1

.

In this paper, we prove the following results: p 1

Theorem 1: Let P ( z )  a 0  a 1 z  ......  a p  1 z such that a

p 1

 a

 apz

 ......  a n z

p

n

be a polynomial of degree n

for some p  1, 2 ,......, n  1 and   0 ,

p

  a n  a n  1  ......  a

 a

p

p 1

 ......  a 1  a 0 .

Then P(z) has at least p zeros in a0 K (2   a n  a 0  K

p (a

 z  K 

n

p 1

)

,

p 1

M

an )

 a

p

where M  2  an  an  a p ,

provided a0 

p  a p  a p 1  M  p 1

  a  

 a0

p 1

 p     M 



p

 a p  a p 1   p 1 

   

p 1

and K<1.

Remark 1: Taking   0 in Theorem 1, we get the following result: Corollary 1: Let P ( z )  a 0  a 1 z  ......  a p  1 z such that a

p 1

 a

p 1

 apz

 ......  a n z

p

n

be a polynomial of degree n

for some p  1, 2 ,......, n  1

p

a n  a n  1  ......  a

p

 a

p 1

 ......  a 1  a 0 .

Then P(z) has at least p zeros in a0 2  an  a0  K

n 1

 z  K

1



an

(a

p M

 a

p

p 1

)

,

p 1

1

where M

1

 an  an  a p ,

provided a0 

p M

1

 a p  a p 1   p 1 

  a  

p 1

 a0



 p    M 1

   

p

 a p  a p 1   p 1 

   

p 1

and K 1  1 .

This result was earlier proved by Roshan Lal et al [4] .

If the coefficients are positive in Theorem 1, we have the following result: Corollary 2: Let P ( z )  a 0  a 1 z  ......  a p  1 z such that a

p 1

 a

p

p 1

 apz

 ......  a n z

p

n

be a polynomial of degree n

for some p  1, 2 ,......, n  1 and   0 ,

  a n  a n  1  ......  a

p

 a

p 1

 ......  a 1  a 0  0 .

Then P(z) has at least p zeros in a0 K 2 {2   a 0  ( K

n 2

 1 ) a n )}

 z  K

2



(a

p M

2

p

 a

p 1

p 1

)

,

where M

2

 2(  a n )  a

p

,

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On The Zeros Of Polynomials... a0 

p M

2

 a p  a p 1   p 1 

  a  

 a0

p 1

 p    M 2



   

p

 a p  a p 1   p 1 

   

p 1

and K 2  1 .

If the coefficients of the polynomial P(z) are complex, we prove the following result: Theorem 2: Let P ( z )  a 0  a 1 z  ......  a p  1 z such that a

p 1

 a

p

p 1

 apz

p

 ......  a n z

n

be a polynomial of degree n

for some p  1, 2 ,......, n  1 and   0 ,

  a n  a n  1  ......  a p  a p  1  ......  a 1  a 0

and for some real  and  , arg a

j

    



, j  0 ,1,......, n .

2

Then P(z) has at least p zeros in z  K

3



p  a p  a p  1 M 3  p 1 

 ,  

where n 1

M

 (   a n )(cos   sin   1 )  a

3

p

cos  



(a

j

 a

j 1

) sin  ,

j  p 1

provided p  a p  a p  1  M 3  p 1 

a0

  m    p    M 3 

   

p

 a p  a p 1   p 1 

   

p 1

Remark 2: Taking   0 in Theorem 2, it reduces to Theorem B. Remark 3: It is easy to see that K 3 <1.

Next, we prove the following result on the zeros of analytic functions : 

Theorem 3: Let f ( z ) 



a jz

j

 0 be analytic in z  K 4 and for some natural number p with

j0

a

p 1

a

 2 

1

,

p

p

  a 0  a 1  a 2  ......  a

 a

p 1

p

 a

p 1

 ...... ,

for some   0 ,and p  a p 1  a p   a p  p 1

a0

  2   a 0  a p  1    

p

 p   a p  a p 1    a   p 1  p  

   

p 1

.

Then f(z) has at least p zeros in z  K

4



a  p 1  p

p 1

a

 a p

p

 .  

Remark 4: Taking   0 , Theorem 3 reduces to the following result: 

Corollary 3: Let f ( z ) 



a jz

j

 0 be analytic in z  K 4 and for some natural number p with

j0

a

p 1

a

p

 2 

1

,

p

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On The Zeros Of Polynomials...   a 0  a 1  a 2  ......  a

 a

p 1

 a

p

 ...... ,

p 1

for some   0 ,and a0

p  a p 1  a p   a p  p 1

  a 0  a p  1    

p

 p   a p  a p 1    a   p 1  p  

   

p 1

.

Then f(z) has at least p zeros in z  K

4

a  p 1  p



 a

p 1

a

 .  

p

p

Cor.3 was earlier proved by Roshan Lal et al [4].

II. LEMMA For the proofs of the above results , we need the following lemma due to Govil and Rahman [1]: Lemma : If a 1 and a 2 are complex numbers such that arg a

j

    



, j  1, 2 , for some real numbers  and  ,

2

then a 1  a 2  ( a 1  a 2 ) cos   ( a 1  a 2 ) sin  .

III.

PROOFS OF THE THEOREMS

3.1 Proof of Theorem 1: Consider the polynomial F ( z )  (1  z ) P ( z )  (1  z )( a 0  a 1 z  ......  a

p 1

z

p 1

 apz

 ......  a n z )

p

n

p 1

 a0 



n

(a

j

 a

j 1

j

)z

 (a

 a

p

p 1

)z

p





j 1

 anz

(a

j

 a

j 1

)z

j

j  p 1 n 1

  (z)   (z) ,

where p 1

 (z)  a0 



(a j  a

j 1

j

)z

j 1

and n

 ( z )  (a

 a

p

)z p 1

p





(a

j

 a

j 1

)  anz

n 1

.

j  p 1

For z  K (<1), we have, by using the hypothesis,  ( z )  a p  a p 1 K

 (a

p

 a

   

p

)K p 1

p

 ( a p  a p 1 ) K

p

 ( a p  a p 1 ) K

p

 ( a p  a p 1 ) K

www.ijceronline.com

p

n



a

j

 a

j 1

K

j

 an K

n 1

j  p 1

 K

p 1

 K

p 1

 K

p 1

 K

p 1

  an K  

a a 2 

    n 1

n p

 a n  a n 1 K

n  p 1





a

j  p 1

n

   a n  a n 1    a n 1  a p

n

   a n  a n 1    a n 1  a p

 an  an  a p



j

 a

j 1

K

n  ( p 1)

   





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Page 175

On The Zeros Of Polynomials...

 (a p

  p ( a p  a p 1 )  a p 1 )   2  an  an  a p   

  p ( a p  a p 1 )    2  an  an  a p     p    2   a n  a n  a p  p     M 

p

p

p 1

2 

 an  an  a p



p 1

p

  a p  a p 1     p 1   

p 1

 a p  a p 1    p 1  

(1)

Also for z  K (<1), p 1

 (z)  a0 



a

j

 a

j

K

j 1

j 1 p 1

 a0  K



(a

j

 a

j 1

)

j 1

 a 0  K ( a p 1  a 0 )  p ( a p  a p 1 )  a0    2  a  a  a n n 

 a0 

p  a p  a p 1  M  p 1

  a  

  a p 1  a 0   p 1 

p

 a0

p 1

   



(2)

Since, by hypothesis a0

p  a p  a p 1   M  p 1

  a  

 p   a p  a p 1  a0      p 1  M   p

p 1

   

p 1

,

it follows from (1) and (2) that  ( z )   ( z ) for z  K .

Hence, by Rouche’s theorem,  ( z ) and  ( z )   ( z ) i.e. F(z) have the same number of zeros in z  K Since the zeros of P(z) are also the zeros of F(z) and since  ( z ) has at least p zeros in z  K ,it follows that P(z) has at least p zeros in z  K .That prove the first part of Theorem 1.

To prove the second part, we show that P(z) has no zero in z 

Let

a0 2  an  a0  an K

n

.

F ( z )  (1  z ) P ( z )  (1  z )( a 0  a 1 z  ......  a

p 1

z

p 1

 apz

p

 ......  a n z ) n

n

 a0 



(a

j

 a

j 1

)z

j

 anz

n 1

j 1

 a0  g (z) ,

where

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Page 176

On The Zeros Of Polynomials... n1

g (z) 



(a

 a

j

 anz

j

)z

j 1

n 1

.

j 1

z  K (<1), we have, by using the hypothesis,

For

n

g (z) 



a

j

 a

j 1

z

a

j

 a

j 1

K

j

 an z

j

 an K

n 1

j 1 n





n 1

j 1

  K  a n  a n 1  

n 1



(a

 a

j

j 1

)  an K

n

j 1

 K   a  a K 2   a  a

  

 K   a n  a n 1    a n 1  a 0  a n K

n

   a n 1  a 0  a n K

n

 

n 1

n

n

 an K

0

n

 

.

Since g(z) is analytic for z  K , g(0)=0, we have, by Schwarz’s lemma,



g (z)  K 2  an  a0  an K

n

z

for z  K .

Hence, for z  K , F (z)  a0  g (z)  a0  g (z)



 a0  K 2  an  a0  an K

n

z

 0

if a0

z 

K (2   a n  a 0  a n K

n

. )

This shows that F(z) and therefore P(z) has no zero in a0

z 

K (2   a n  a 0  a n K

n

. )

That proves Theorem 1 completely.

Proof of Theorem 2: Consider the polynomial F ( z )  (1  z ) P ( z )  (1  z )( a 0  a 1 z  ......  a

p 1

z

p 1

 apz

p

p 1

 a0 



n

n

(a

j

 a

)z j 1

j 1

 anz

 ......  a n z )

j

 (a

p

 a

)z p 1

p





(a

j

 a

j 1

)z

j

j  p 1

n 1

  (z)   (z) ,

where p 1

 (z)  a0 



(a j  a

j 1

)z

j

j 1

and

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Page 177

On The Zeros Of Polynomials... n

 ( z )  (a

 a

p





p

)z p 1

(a

j

 a

j 1

n 1

)  anz

..

j  p 1

For z  K 3 (<1), we have, by using the hypothesis,  ( z )  a p  a p 1 K 3

 a

 a

 a

p

p

 a

 a

 a

p

p

K

p 1

p

K

p 1

 a p  a p 1 K 3

p

a

 a

j

j 1

K

j 3

 an K

   

n 1 3

j  p 1

3

3

p 1

 K

3



p 1

 K

3

n

p 1

 K

3

p

K

p 1

   

p

3

 K3

p 1

  an K  

n p 3

 a n  a n 1 K

p 1

K

3



(a

 a

j

a

j

 a

j 1

   

a

j

 a

j 1

   

j  p 1 n 1

 j  p 1

j 1

K

n  ( p 1) 3

   

p 1

 a

p

) cos   ( a

 a

p 1

p

) sin 



a n )(cos   sin   1 )  a p cos 



) sin 

j 1

 a

 a n  {(   a n )  a n  1 } cos  

(  

n 1



j

n 1



 ......  ( a n  1  a n  2 ) cos   ( a n  1  a n  2 ) sin  p

a

j  p 1

  a n    a n  a n 1     

{(   a n )  a n  1 } sin   ( a

 a p  a p 1 K 3





3

  a n    a n  a n 1     



n 1

n  p 1

j  p 1

p

 a p  a p 1 K 3

 a

p

 a

p

p 1

K

3

M

3

 p a p  a p 1   M p 1 3 

p 1

 a p  a p 1   p 1 

   

 p     M 3

   

 a0 

p  a p  a p  1 M 3  p 1 

   

p

 p a p  a p 1    M p 1 3 

   

p 1

M

3

p 1

 m   

 a0  K 3m 

(3)

Also , for z  K 3 , we have, by using the lemma and the hypothesis, j

p 1

 (z)  a0 



a

j

 a

j 1

z

a

j

 a

j 1

K

j 1

p 1

 a0 



j 3

j 1 p 1

 a0  K

3



a

j

 a

j 1

j 1

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On The Zeros Of Polynomials...  a 0  K 3 ( a 1  a 0 ) cos   ( a 1  a 0 ) sin   .......











 ( a p  1  a p  2 ) cos   ( a p  1  a p  2 ) sin   a0  K

3



(a



 a 0 ) cos  

p 1

(a

 a

j

 a0  K 3m 



) sin 

j 1

(4)

Thus, for z  K 3 , we have from (3) and (4),  ( z )   ( z ) .

Since  ( z ) and  ( z ) are analytic for z  K 3 , it follows by Rouche’s theorem that  ( z ) and  ( z )   ( z ) i.e. F(z) have the same number of zeros in z  K 3 . But the zeros of P(z) are also the zeros of

F(z). Therefore, we conclude that P(z) has at least p zeros in z  K , as the same is true of  ( z ) . That proves Theorem 2.

Proof of Theorem 3: Consider the function F ( z )  ( z  1) f ( z )  ( z  1 )( a 0  a 1 z  a 2 z

 ......)

2

p 1



 a0 



(a

 a j )z

j 1

 (a

j

 a p )z

p 1



p

j 1



(a

 a j )z

j 1

j

j  p 1

  ( z )   ( z ),

where p 1



 (z)  a0 

(a

 a j )z

j 1

j

,

j 1 

 ( z )  (a

 a p )z

p 1





p

(a

 a j )z

j 1

j

.

j  p 1

a

For z  K 4 ( K 4 <1, by hypothesis for  ( z )  a p 1  a p K

 a

 a

p 1

K

p

 a p 1  a p K  a

p 1

 a

p

p 4

p 4

p 4

 K

4

4

 K

p 1 4

 a p )( ( p 1 

p

   

   

p 1

 K

 p   a p 1  a p     a   p 1  p  

   

p 1

p 1

a

p

a

j 1

1

 2 





), we have

p  aj K

j  ( p 1) 4

j  p 1





a

j  p 1

j 1

 a

j

   

   

ap

p 1

a

 a p

p

  a p )  ( )(   p  1

p 1

a

 a

p

p

 ) a 

p

p 1

.

(5)

and p 1

 (z)  a0 



a

j 1

 a

j

K

j 4

j 1

 a0 



a 0  a 1  a 1  a 2  ......  a p  2  a p  1

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K 4

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Page 179

On The Zeros Of Polynomials...  a0 



  a 0  a 1    a 1  a 2  .....  a p  2  a p  1

K 4

 a0 



  a 0  a 1    a 1  a 2  .....  a

K 4

 a0 



  a 0  a 1    a 1  a 2  .....  a

 a0 



2  a0  a

 a0 

p  a p 1  a  a p  p 1

p 1

p

p2

p2

 a

 a

p 1

p 1

K 4

K 4

  2   a 0  a  

p 1



(6)

Since, by hypothesis, a0

p  a p 1  a p   a p  p 1

  2   a 0  a p  1    

p

 p   a p  a p 1    a   p 1  p  

   

p 1

,

it follows from (5) and (6) that  ( z )   ( z ) for z  K 4 .

Since  ( z ) and  ( z ) are analytic for z  K 3 , it follows, by Rouche’s theorem, that  ( z ) and  ( z )   ( z ) i.e. F(z) have the same number of zeros in z  K 3 . But the zeros of P(z) are also the zeros of

F(z). Therefore, we conclude that P(z) has at least p zeros in z  K , as the same is true of  ( z ) . That proves Theorem 3.

REFERENCES [1] [2] [3] [4]

N. K. Govil and Q. I. Rahman, On the Enestrom-Kakeya Theorem, Tohoku Math. J.20 (1968), 126136.

V.K. Jain, On the zeros of a polynomial, Proc.Indian Acad. Sci. Math. Sci.119 (1) ,2009, 37-43.

M. Marden, Geometry of Polynomials, Math. Surveys No. 3, Amer. Math. Soc. Providence, RI, 1966.

Roshan Lal, Susheel Kumar and Sunil Hans, On the zeros of polynomials and analytic functions, Annales Universitatis Mariae Curie –Sklodowska Lublin Polonia, Vol.LXV, No. 1, 2011, Sectio A, 97-108.

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