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International Journal of Scientific and Research Publications, Volume 3, Issue 2, February 2013 767
ISSN 2250-3153
a
If
equation
and
n− k −1
>
δ =
1
a n
a
a
n−k
(i.e. λ > 1), then P(z) has all its zeros in the disk
K
( λ −1)
γ
1
=
a
n
k + 1
an
− k
− δ K
,
1
k
− γ
1
= 0
n
z ≤ K 1
, where
K1
n
an
an
k
a
a ∑ − 1
+ ( λ −1)
−
}(cosα
+ sinα)
− µ
0
(cosα
− sinα
+ 1) + 2
0
+ 2sin
j=
1
[{ α
− k
>
n−k
+ 1
If
equation
and
< λ z ≤ K
2
), then P(z) has all its zeros in the disk
(i.e. 0 < 1
K
(1 − λ)
γ
2
=
a
n
k
an
− k
− δ K
,
γ
k −1
2
−
2
=
n
0
a
, where
K
2
[{ an
(1 ) an
k
}(cos sin ) a0
(cos sin 1) 2 a0
2 sin ∑ − + − λ
−
α + α − µ α − α + + + α a
j=
1
δ
2
=
a
n 1
j
]
.
is the greatest positive root of the
a
j
]
.
is the greatest positive root of the
II. LEMMAS
For the proofs of the above results , we need the following result:
Lemma 1: Let P(z)=
n
∑
j=
a j z
π
arg a j
− β ≤ α ≤ , j = 0,1,... n,
for
2
ta
j
− a
≤
j
0
be a polynomial of degree n with complex coefficients such that
some real
β.
Then for some t>0,
[ t a − a ] α + [ t a + a ] sin .
cos
j j−1
j−1 j j−1
α
The proof of lemma 1 follows from a lemma due to Govil and Rahman [2].
Proof of Theorem 1: We first prove that
K
1 >1 and
K
2 >1.
Let
f ( K = K − δ K − γ
1
)
k + 1 k
1
.
To prove
K
1 >1, it suffices to prove that
f 1
(1) < 0.
If
(a)
(b)
a
n− k −1 > an−k
a
a
,
1
then one of the following four cases happens:
n−k
+ 1
≥ an−k
−1
> an−k
>
0
and λ > 1.
n− k + 1
≥ an−k
−1
≥ 0 > an−k
and ≤ 0
λ .
III. PROOFS OF THE THEOREMS
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