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International Journal of Scientific and Research Publications, Volume 3, Issue 2, February 2013 771
ISSN 2250-3153
k + 1
z − γ
1
> δ
1
z
k
.
Hence, it follows that all the zeros of F(z) whose modulus is greater than 1 lie in the disk
root of the equation
K
k + 1
− δ K
1
k
− γ
1
= 0
z ≤ K 1
, where
K1
is the greatest positive
.
As in the proof of Theorem 1, it can be shown that
K > 1 1 , so that the zeros of F(z) whose modulus is less than or equal to 1 are
z ≤ K
already contained in the disk 1
z ≤ K
. Therefore, all the zeros of F(z) and hence P(z) lie in 1
.
α n
α α n
α
Now,, suppose that
,
− k
>
n−k
+1
then
,
− k
>
n−k
−1
and we have
n 1
n−k
+ 1 n
F( z)
= −an
z − (1 − λ)
α
n−k
z − ρz
+ ( ρ + α
n
− α
n−1
) z
n−k
+ 1
n−k
+ ( α − λα ) z + ( α − α ) z
For
if
F
z > 1
,
+ ( α
+ n
n−k
+ 1
n−k
−1
− α
n−k
n−k
−2
) z
n−k
−1
n
n−k
n−k
−1
+ ...... + ( α − µα ) z
j
+ ( µ −1)
α
0
z + α
0
+ i∑
( β
j
− β
j−1
) z + iβ
0.
j=
1
1
0
+ ......
n+
1 n−k
+ 1 n
1
( z)
≥ an
z + (1 − λ)
α
n−k
z − z [ ρ + ( ρ + α
n
− α
n−1
) + ...... + ( α
n−k
+ 1
− λα
n−k
)
k −1
+ ( α
n−k
− α
(1 − µ ) α
0
+
n−1
z
n−k
−1
)
z
1
k
+ ( α
α
0
+ ] − z
n
z
n
n−k
−1
[( β − β
n
− α
n−k
−2
n−1
)
z
1
k + 1
1
+ ...... + ( α1
− µα
0
)
z
1
) + ...... + ( β1
− β
0
)
z
n+
1
n
> an
z + (
n−k
n
n−k
0 0
>0
n−1
n−k
+ 1
1 − λ ) α z − z [2ρ
+ α + (1 − λ)
α − µ ( α + α )
+ 2α 0
+ β
n
− β
0
+ β
0
z
k
+ γ
But this inequality holds if
z
k
− γ
2
2
> δ z
2
2
> δ z
k −1
k −1
]
.
.
2
Hence, it follows that all the zeros of F(z) whose modulus is greater than 1 lie in the disk , where
K
2 is the greatest positive
root of the equation
K
k
− δ
k −1
2
K − γ
1
=
0
.
As in the proof of Theorem 1, it can be shown that
K > 1 2 .Thus the zeros of F(z) whose modulus is less than or equal to 1 are
z ≤ K
already contained in the disk
2
. Therefore, all the zeros of F(z) and hence P(z) lie in
2
and the proof of Theorem 2 is
complete.
Proof of Theorem 3: Consider the polynomial
F( z)
= (1 − z)
P(
z)
= (1 − z)(
a
= −a
n
z
n+
1
n
z
n
+ ( a
+ a
n
n−1
− a
z
n−1
n−1
) z
+ ...... + a z + a )
n
1
+ ...... + ( a
0
n−k
+ 1
− a
n−k
) z
n−k
+ 1
+ ( a
n−k
+
− a
β
z
0
n
n−k
−1
n−1
]
z ≤ K
) z
z ≤ K
n−k
z
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