TITLE MARCH 2012 - Pakistan Academy of Sciences
TITLE MARCH 2012 - Pakistan Academy of Sciences
TITLE MARCH 2012 - Pakistan Academy of Sciences
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Sukhwinder Singh Billing 5<br />
Corollary 3.1. Let be a real number with<br />
zf ()<br />
z<br />
0 . If f , 0, zE<br />
, satisfies the<br />
f()<br />
z<br />
differential subordination<br />
zf ( z) zf ( z)<br />
<br />
(1 ) 1<br />
<br />
f ( z) f ( z)<br />
<br />
1 (1 2 ) z 2 (1 )<br />
z<br />
<br />
,<br />
1 z (1 z)(1 (1 2 ) z)<br />
can vary over the portion <strong>of</strong> the plane right to the<br />
curve h () 1<br />
z for the same conclusion. Thus our<br />
result extends the region <strong>of</strong> variability <strong>of</strong> this<br />
operator for the same implication and the region<br />
bounded by the dashing line and the curve is the<br />
claimed extension as shown in Fig. 1.<br />
then<br />
f<br />
* ( )<br />
.<br />
1 3<br />
Remark 3.2. For and , Corollary<br />
2 4<br />
3.1 reduces to the following result.<br />
zf ()<br />
z<br />
If f , 0, zE<br />
, satisfies the<br />
f()<br />
z<br />
condition<br />
zf ( z) zf ( z) 2 z z<br />
1 h1<br />
( z),<br />
f ( z) f ( z) 1 z (1 z)(2 z)<br />
then<br />
f * (3/ 4) .<br />
Fig. 1.<br />
Substituting the same values <strong>of</strong> and in<br />
the result <strong>of</strong> Fukui [1] stated in Theorem 1.2, we<br />
obtain the following result:<br />
If f <br />
, satisfies the condition<br />
zf ( z) zf ( z) 4<br />
1 , z E,<br />
f ( z) f ( z) 3<br />
then<br />
f * (3/ 4) .<br />
To compare both the results, we plot h ( ) 1<br />
4<br />
and the line ( z)<br />
in Fig. 1.<br />
3<br />
We see that according to the result <strong>of</strong> Fukui<br />
[1], for the starlikeness <strong>of</strong> order 3/ 4 <strong>of</strong> f()<br />
z ,<br />
zf ( z) zf ( z)<br />
the differential operator 1<br />
can<br />
f ( z) f ( z)<br />
vary in the complex plane on the right side <strong>of</strong> the<br />
4<br />
line ( z)<br />
shown with dashes in Fig. 1<br />
3<br />
whereas according to our result, the same operator<br />
Remark 3.3. When we select<br />
1 (1 2 )<br />
z<br />
qz ( ) , 0 1. Then<br />
1<br />
z<br />
zq<br />
( z) 1 z zq<br />
( z)<br />
<br />
1 , i.e. 1 0,<br />
q<br />
<br />
( z) 1 z q( z)<br />
<br />
and<br />
zq( z) 1<br />
1z<br />
1 2 qz ( ) <br />
q( z) 1<br />
z<br />
1 (1 2 ) z 1<br />
2 .<br />
1<br />
z <br />
Therefore for 0 <br />
1, we have<br />
zq( z) 1<br />
<br />
1 2 qz ( ) <br />
0.<br />
q()<br />
z