On coefficient problems for functions starlike with respect to symmetric points

Lemma 1 ([5]) Let xðzÞ ¼ c1z þ c2z2 þ    be a Schwarz function. Then, for any real numbers l and m such that ðl; mÞ 2 jlj  1 2 ; 1  m  1   [ 1 2  jlj  2 ; 4 27 jlj þ 1 ð Þ3 jlj þ 1 ð

Lemma 2 ( [1]) Let xðzÞ ¼ c1z þ c2z2 þ    be a Schwarz function. Then, jc3j  1  jc1j2  jc2j2 1 þ jc1j ; jc4j  1  jc1j2  jc2j2 ; jc5j  1  jc1j2  jc2j2  jc3j2 1 þ jc1j : The above lemma immediately results in the following fact.

Lemma 3 Let xðzÞ ¼ c1z þ c2z2 þ    be a Schwarz function. Then jc1c3  c2 2j  1  jc1j2 : We also need the results obtained by Efraimidis.

Lemma 4 ([2]) Let xðzÞ ¼ c1z þ c2z2 þ    be a Schwarz function and k 2 C. Then c4 þ ð1 þ kÞc1c3 þ c2 2 þ ð1 þ 2kÞc2 1c2 þ kc4 1    max 1; jkj f g ð1:4Þ and On coefficient problems for functions... Page 3 of 12 17

Lemma 5 If x 2 B0 is of the form (1.2) and l 2 C, jlj  1, then c5 þ ð1 þ lÞc1c4 þ ð1 þ lÞc2c3 þ 3lc1c2 2 þ 1 þ l þ l2   c2 1c3  þ2lð1 þ lÞc3 1c2 þ l2c5 1   1 : ð1:6Þ 2 Coefficient bounds

Theorem 1 If f 2 S SðezÞ is of the form (1.1), then ja4j  1 4 and ja5j  1 4 : The bounds are sharp.

Lemma 1 with l ¼ 3 2 and m ¼ 5 12 applied to ja4j ¼ 1 4 c3 þ 3 2c1c2 þ 5 12c3 1  ; results in the first inequality. To prove the second inequality, we can write 17 Page 4 of 12 P. Zaprawa

Theorem 2 If f 2 S SðezÞ is of the form (1.1), then jc1j  1 4 ; jc2j  1 4 ; jc3j  1 8 ; jc4j  1 8 ;

Lemma 1 with l ¼ 1 2 and m ¼ 1 12. Observe that c4 þ 1 2c1c3 þ 1 2c2 2 þ 1 4c2 1c2 þ 1 48c4 1   ¼ 1

Theorem 3 If f 2 S SðezÞ is of the form (1.1), then the following sharp bounds hold ja4  a2a3j  1 4 and ja5  a2 3j  1 4 : Let us turn to Hankel determinants for the class S SðezÞ. The first result is easy to obtain.

Theorem 4 If f 2 S SðezÞ is of the form (1.1), then jH2;2j  1 4 :

Theorem 5 If f 2 S SðezÞ is of the form (1.1), the following sharp bound holds jH2;3j  1 8 :

Theorem 6 If f 2 S SðezÞ is of the form (1.1), then jH3;1j  13 128 :