1. International Journal of Modern Engineering Research (IJMER)
www.ijmer.com Vol.3, Issue.1, Jan-Feb. 2013 pp-240-249 ISSN: 2249-6645
On The Number of Zeros of a Polynomial inside the Unit Disk
M. H. Gulzar
Department of Mathematics University of Kashmir, Srinagar 190006
Abstract: In this paper we find an upper bound for the number of zeros of a polynomial inside the unit disk, when the
coefficients of the polynomial or their real and imaginary parts are restricted to certain conditions.
Mathematics Subject Classification: 30C10, 30C15
Key-words and phrases: Polynomial, Unit disk, Zero.
I. Introduction and Statement of Results
Regarding the number of zeros of a polynomial inside the unit disk, the following results were recently proved by
M. H. Gulzar [2]:
n
Theorem A: Let P (z) = a z
j 0
j
j
be a polynomial of degree n such that for some   0 ,
  a n  a n1  ......  a1  a0 .
a0
Then the number of zeros of P (z) in  z   ,0    1 does not exceed
M1
1 2  a n  a n  a0  a0
log ,
1 a0
log

Where M 1  2  a n  a n  a0 .
n
TheoremB: Let P (z) = a z
j 0
j
j
be a polynomial of degree n with complex coefficients .If
Re a j   j , Im a j   j , j  0,1,..., n, and for some   0,
   n   n1  ...  1   0 ,
a0
then the number of zeros of P(z) in M
 z   ,0    1, does not exceed
2
n
2    n   n   0   0  2  j
1 j 0
log ,
1 a0
log

n
Where M 2  2    n   n   0   0  2  j .
j 1
n
Theorem C: Let P(z)= a z
j 0
j
j
be a polynomial of degree n with complex coefficients .If
Re a j   j , Im a j   j , j  0,1,..., n, and for some   0,
   n   n1  ...  1   0 ,
a0
 z   ,0    1,
then the number of zeros of P(z) in M does not exceed
3
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2. International Journal of Modern Engineering Research (IJMER)
www.ijmer.com Vol.3, Issue.1, Jan-Feb. 2013 pp-240-249 ISSN: 2249-6645
n
2    n   n   0   0  2  j
1 j 0
log ,
1 a0
log

n
Where M 3  2    n   n   0   0  2  j .
j 1
n
Theorem D. Let P (z) = a z
j 0
j
j
be a polynomial of degree n with complex coefficients such that

arg a j      , j  0,1,...n, for some real 
2
and
  a n  a n1  ...  a1  a0 ,for some   0.
a0
Then the number of zeros of P(z) in  z   ,0    1, does not exceed
M4
n 1
(   a n )(cos   sin   1)  a 0 (cos   sin   1)  2 sin   a j
1 j 1
log ,
1 a0
log

where
n 1
M 4  (   a n )(cos   sin   1)  a 0 (cos   sin  )  2 sin   a j
j 1
In this paper, we prove certain generalizations of the above results. In fact, we prove the following :
n
Theorem 1: Let P( z )  a
j 0
j z j be a polynomial of degree n with Re( a j )   j and Im(a j )   j , j=0,1,2,……,n. If
for some real numbers  ,   0 , 1  k  n,  nk  0,  nk 1   nk ,
   n   n1  ... nk 1   nk   nk 1  ...  1   0 ,
a0
then the number of zeros of P(z) in  z   ,0    1 , does not exceed
M5
n
2    n   n  (  1) n  k    1  n  k   0   0  2  j
1 j 0
log ,
1 a0
log

n
where M 5  2    n   n  (  1) n  k    1  n  k   0   0  2  j ,
j 1
a0
and if  nk   nk 1 , then the number of zeros of P(z) in  z   ,0    1 , does not exceed
M6
n
2    n   n  (1   ) n  k  1    n k   0   0  2  j
1 j 0
log ,
1 a0
log

n
where M 6  2    n   n  (1   ) n  k  1    n  k   0   0  2  j .
j 1
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3. International Journal of Modern Engineering Research (IJMER)
www.ijmer.com Vol.3, Issue.1, Jan-Feb. 2013 pp-240-249 ISSN: 2249-6645
Remark 1: Taking   1 , Theorem 1 reduces to Theorem B.
Remark 2: If a j are real i.e.  j  0 for all j , Theorem 1 gives the following result which reduces to Theorem A by
taking   1:
n
Theorem 2: Let P( z )   a j z j be a polynomial of degree n .If for some real numbers  ,  0,
j 0
1  k  n, ank  0, ank 1  ank ,
  an  an1  ...ank 1  ank  ank 1  ...  a1  a0 ,
a0
then the number of zeros of P(z) in  z   ,0    1 , does not exceed
M7
1 2   a n  a n  (  1)a n k    1 a n k  a0  a0
log ,
1 a0
log

where M 7  2  a n  a n  (  1)a n  k    1 a nk  a0 ,
a0
and if ank  ank 1 , then the number of zeros of P(z) in  z   ,0    1 , does not exceed
M8
1 2   a n  a n  (1   )a n k  1   a n k  a0  a0
log ,
1 a0
log

where M 8  2  a n  a n  (1   )a n  k  1   a n  k  a0 .
Applying Theorem 1 to the polynomial –iP(z), we get the following result, which reduces to Theorem C by taking  1:
n
Theorem 3: Let P( z )  a
j 0
j z j be a polynomial of degree n with Re( a j )   j and Im(a j )   j , j=0,1,2,……,n. If
for some real numbers  ,   0 , 1  k  n,  nk  0,  n k 1   n k ,
   n   n1  ... nk 1   nk   nk 1  ...  1   0 ,
a0
then the number of zeros of P(z) in  z   ,0    1 , does not exceed
M9
n
2    n   n  (  1)  n  k    1  n  k   0   0  2  j
1 j 0
log ,
1 a0
log

n
where M 9  2    n   n  (  1)  n  k    1  n  k   0   0  2  j ,
j 1
a0
and if  n k   n k 1 , then the number of zeros of P(z) in  z   ,0    1 , does not exceed
M 10
n
2    n   n  (1   )  n  k  1    n  k   0   0  2  j
1 j 0
log ,
1 a0
log

n
where M 10  2    n   n  (1   )  n  k  1    n  k   0   0  2  j .
j 1
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4. International Journal of Modern Engineering Research (IJMER)
www.ijmer.com Vol.3, Issue.1, Jan-Feb. 2013 pp-240-249 ISSN: 2249-6645
n
Theorem 4: Let P( z )  a
j 0
j z j be a polynomial of degree n .If for some real numbers   0 ,   0 ,
1  k  n, ank  0,
  an  an1  ...  ank 1   ank  ank 1  ...  a1  a0 ,

and for some real  , arg a j      , j  0,1,......, n and an k 1  an k , i.e.   1 ,then the number of zeros of
2
a0
P(z) in  z   ,0    1 , does not exceed
M 11
[(   a n (cos   sin   1)  a n  k (cos   sin    cos    sin     1)
n 1
1
 a 0 (cos   sin   1)  2 sin  a
j 1, j  n  k
j ]
log
1 a0
log

where M11  (   an (cos   sin   1)  an  k (cos   sin    cos    sin     1)
n 1
 a 0 (cos   sin  )  2 sin  a
j 1, j  n  k
j
a0
and if an k  a n k 1 , i.e.   1 , then the number of zeros of P(z) in  z   ,0    1 , does not exceed
M 12
[(   a n (cos   sin   1)  a n  k (cos   sin    cos    sin   1   )
n 1
1
 a 0 (cos   sin   1)  2 sin  a
j 1, j  n  k
j ]
log Where
1 a0
log

M 12  (   a n (cos   sin   1)  a nk (cos   sin    cos    sin   1   )
n 1
 a 0 (cos   sin  )  2 sin  a
j 1, j  n  k
j .
Remark 4: Taking   1 , Theorem 4 reduces to Theorem D.
II. Lemmas
For the proofs of the above results, we need the following results:
n
Lemma 1:. Let P(z)= a z
j 0
j
j
be a polynomial of degree n with complex coefficients such that

arg a j      , j  0,1,...n, for some real  ,then for some t>0,
2
ta j  a j 1   
 t a j  a j 1 cos   t a j  a j1 sin  . 
The proof of lemma 1 follows from a lemma due to Govil and Rahman [1].
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5. International Journal of Modern Engineering Research (IJMER)
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Lemma 2.If p(z) is regular ,p(0)  0 and p( z )  M in z  1, then the number of zeros of p(z) in z   ,0    1, does
1 M
not exceed log (see[4],p171).
1 p(0)
log

III. Proofs of Theorems
Proof of Theorem 1: Consider the polynomial
F ( z)  (1  z) P( z)
 (1  z )(a n z n  a n1 z n 1  ......  a1 z  a0 )
 a n z n1  (a n  a n1 ) z n  ......  (a nk 1  a n k ) z n k 1  (a nk  a nk 1 ) z n k
 (ank 1  ank 2 ) z nk 1  ......  (a1  a0 ) z  a0
 ( n  i n ) z n 1  ( n   n 1 ) z n  ......  ( n  k 1   n k ) z n  k 1
 ( n  k   n  k 1 ) z n  k  ( n k 1   n  k  2 ) z n  k 1  ......  ( 1   0 ) z   0
n
 i  (  j   j 1 ) z j  i 0
j 1
If  nk 1   nk , then
F ( z)  ( n  i n ) z n1  z n  (    n   n1 ) z n  ......  ( nk 1   nk ) z nk 1
 ( n k   n k 1 ) z n k  (  1) n k z n k  ( n k 1   n k 2 ) z n k 1  ......
n
 (1   0 ) z   0  i  (  j   j 1 ) z j  i 0 .
j 1
For z  1 ,
F ( z )   n       n   n1  ......   nk 1   nk +  nk   nk 1    1  nk
n
  nk 1   nk 2  ......  1   0   0  2  j
j 0
n
 2    n   n  (  1) nk    1  nk   0   0  2  j
j 0
Hence by Lemma 2, the number of zeros of F(z) in z   ,0    1, does not exceed
n
2    n   n  (  1) n  k    1  n k   0   0  2  j
1 j 0
log .
1 a0
log

On the other hand, let
n 1
Q (z)=  an z  (an  an1 ) z n  ......  (ank 1  ank ) z nk 1  (ank  ank 1 ) z nk
 (ank 1  ank 2 ) z nk 1  ......  (a1  a0 ) z
For z  1 ,
Q( z )   n       n   n1  ......   nk 1   nk   nk   nk 1    1  nk
n
   k 1   n  k  2  ......   1   0   0  2  j
j 1
n
 2    n   n  (  1)    1  n  k   0   0  2  j  M 5 .
j 1
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6. International Journal of Modern Engineering Research (IJMER)
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Since Q(0)=0, we have, by Rouche’s theorem,
Q( z )  M 5 z , for z  1 .
Thus
F ( z )  a0  Q( z )
 a 0  Q( z )
 a0  M 5 z
0
a0
if z  .
M5
a0
This shows that F(z) has no zero in z  . Consequently it follows that the number of zeros of F(z) and hence P(z) in
M5
a0
 z   ,0    1 , does not exceed
M5
n
2    n   n  (  1) n  k    1  n k   0   0  2  j
1 j 0
log
1 a0
log

If  nk   nk 1 , then
F ( z)  ( n  i n ) z n1  z n  (    n   n1 ) z n  ......  ( nk 1   nk ) z nk 1
 ( n k   n k 1 ) z n k  (1   ) n k z n k   ( n k 1   n k 2 ) z n k 1  ......
n
 ( 1   0 ) z   0  i  (  j   j 1 ) z j  i 0 .
j 1
For z  1 ,
F ( z )   n       n   n1  ......   nk 1   nk +  nk   nk 1  1    nk
n
  nk 1   nk 2  ......  1   0   0  2  j
j 0
n
 2    n   n  (1   ) nk  1    nk   0   0  2  j
j 0
Hence by Lemma 2, the number of zeros of F(z) in z   ,0    1, does not exceed
n
2    n   n  (1   ) n  k  1    n k   0   0  2  j
1 j 0
log .
1 a0
log

On the other hand, let
n 1
Q(z)=  an z  (an  an1 ) z n  ......  (ank 1  ank ) z nk 1  (ank  ank 1 ) z nk
 (ank 1  ank 2 ) z nk 1  ......  (a1  a0 ) z
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 ( n  i n ) z n 1  z n  (    n   n 1 ) z n  ......  ( n  k 1   n  k ) z n k 1
 ( n k   n  k 1 ) z n  k  (1   )a n  k z n k 1  ( n  k 1   n k  2 ) z n k 1  ......
n
 ( 1   0 ) z  i  (  j   j 1 ) z j
j 1
For z  1 ,
Q( z )   n       n   n1  ......   nk 1   nk   nk   nk 1  1    nk
n
  k 1   nk 2  ......   1   0   0  2  j
j 1
n
 2    n   n  (1   )  1    n  k   0   0  2  j  M 6 .
j 1
Since Q(0)=0, we have, by Rouche’s theorem,
Q( z )  M z , for z  1 .
Thus
F ( z )  a0  Q( z )
 a 0  Q( z )
 a0  M 6 z
0
a0
if z  .
M6
a0
This shows that F(z) has no zero in z  . Consequently it follows that the number of zeros of F(z) and hence P(z) in
M6
a0
 z   ,0    1, does not exceed
M6
n
2    n   n  (1   ) n  k  1    n k   0   0  2  j
1 j 0
log .
1 a0
log

That proves Theorem 1.
Proof of Theorem 4: Consider the polynomial
F ( z)  (1  z) P( z)
 (1  z )(a n z n  a n1 z n 1  ......  a1 z  a0 )
 a n z n1  (a n  a n1 ) z n  ......  (a nk 1  a n k ) z n k 1  (a nk  a nk 1 ) z n k
 (ank 1  ank 2 ) z nk 1  ......  (a1  a0 ) z  a0
If an k 1  an k , i.e.   1 , then
F ( z )  a n z n 1
 z n  (   an  an1 ) z n  ......  (ank 1  ank ) z nk 1
 (ank  ank 1 ) z nk  (  1)ank z nk  (ank 1  ank 2 ) z nk 1  ......
 (a1  a0 ) z  a0
so that for z  1 , we have by using Lemma 1,
F ( z )  an      an  an1  ......  ank 1  ank  ank  ank 1
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   1 ank  ank 1  ank 2  ......  a1  a0  a0
 an    (   an  an1 ) cos   (   an  an1 ) sin   ......
 ( a n  k 1  a n  k ) cos   ( a n  k 1  a n  k ) sin   (  1) a n  k
 ( a n  k  a n  k 1 ) cos   ( a n  k  a n  k 1 ) sin 
 ( a n  k 1  a n  k  2 ) cos   ( a n  k 1  a n  k  2 ) sin   ......
 ( a1  a 0 ) cos   ( a1  a 0 ) sin   a 0
 (   a n )(cos   sin   1)  a nk (cos   sin    cos    sin     1)
n 1
 a0 (cos   sin   1)  2 sin  a
j 1, j  n  k
j
Hence, by Lemma 2, the number of zeros of F(z) in z   ,0    1, does not exceed
[(   a n (cos   sin   1)  a n  k (cos   sin    cos    sin     1)
n 1
1
 a 0 (cos   sin   1)  2 sin  a
j 1, j  n  k
j ]
log On the other hand,
1 a0
log

let
n 1
Q(z)=  an z  (an  an1 ) z n  ......  (ank 1  ank ) z nk 1  (ank  ank 1 ) z nk
 (ank 1  ank 2 ) z nk 1  ......  (a1  a0 ) z
For z  1 ,
Q(z )  (   a n )(cos   sin   1)  a nk (cos   sin    cos    sin     1)
n 1
 a0 (cos   sin  )  2 sin  a
j 1, j  n  k
j
 M 11 .
Since Q(0)=0 , we have , by Rouche’s Theorem,
Q( z )  M 11 z , for z  1 .
Thus, for z  1 ,
F ( z )  a0  Q( z )
 a 0  Q( z )
 a 0  M 11 z
0
a0
if z  .
M 11
a0
This shows that F(z) has all its zeros z with z  1 in z  .
M 11
a0
Thus, the number of zeros of F(z) and hence P(z) in  z   ,0    1 , does not exceed
M 11
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[(   a n (cos   sin   1)  a n  k (cos   sin    cos    sin     1)
n 1
1
 a 0 (cos   sin   1)  2 sin  a
j 1, j  n  k
j ]
log
1 a0
log

If an k  a n k 1 , i.e.   1 , then
F ( z)  an z n1  z n  (   an  an1 ) z n  ......  (ank 1  ank ) z nk 1
 (ank  ank 1 ) z nk  (1   )ank z nk 1  (ank 1  ank 2 ) z nk 1  ......
 (a1  a0 ) z  a0
so that for z  1 , we have by Lemma 1,
F ( z )  an      an  an1  ......  ank 1  ank  ank  ank 1
 1   ank  ank 1  ank 2  ......  a1  a0  a0
 an    (   an  an1 ) cos   (   an  an1 ) sin   ......
 ( a n k 1   a n  k ) cos   ( a n  k 1   a n  k ) sin   1   a n k
 ( a n k  a n  k 1 ) cos   ( a n  k  a n k 1 ) sin 
 ( a n k 1  a n  k  2 ) cos   ( a n  k 1  a n  k  2 ) sin   ......
 ( a1  a 0 ) cos   ( a1  a 0 ) sin   a 0
 (   a n )(cos   sin   1)  a nk (cos   sin    cos    sin   1   )
n 1
 a0 (cos   sin   1)  2 sin  a
j 1, j  n  k
j
Hence , by Lemma 2, the number of zeros of F(z) in z   ,0    1, does not exceed
[(   an (cos   sin   1)  an  k (cos   sin    cos    sin   1   )
n 1
1
log
 a0 (cos   sin   1)  2 sin  a
j 1, j  n  k
j ]
1
log a0

On the other hand, let
n 1
Q(z)=  an z  (an  an1 ) z n  ......  (ank 1  ank ) z nk 1  (ank  ank 1 ) z nk
 (ank 1  ank 2 ) z nk 1  ......  (a1  a0 ) z
For z  1 , by using Lemma 1,
Q(z )  (   a n )(cos   sin   1)  a n k (cos   sin    cos    sin   1   )
n 1
 a0 (cos   sin  )  2 sin  a
j 1, j  n  k
j
 M 12 .
Since Q(0)=0 , we have , by Rouche’s Theorem,
Q( z )  M 12 z , for z  1 .
Thus, for z  1 ,
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F ( z )  a0  Q( z )
 a 0  Q( z )
 a 0  M 12 z
0
a0
if z  .
M 12
a0
This shows that F(z) has all its zeros z with z  1 in z  .
M 12
a0
Thus, the number of zeros of F(z) and hence P(z) in  z   ,0    1 , does not exceed
M 12
[(   an (cos   sin   1)  an  k (cos   sin    cos    sin   1   )
n 1
1
 a0 (cos   sin   1)  2 sin  a
j 1, j  n  k
j ]
log .
1 a0
log

That proves Theorem 4.
References
[1] N.K.Govil and Q.I.Rahman, On the Enestrom-Kakeya Theorem, Tohoku Math.J.20 (1968), 126-136.
[2] M. H. Gulzar, on the number of Zeros of a polynomial in a prescribed Region, Research Journal of Pure Algebra-2(2), 2012, 1-11.
[3] E C. Titchmarsh, The Theory of Functions, 2nd edition, Oxford University Press, London, 1939.
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