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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

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