设f(z)=h(z)+g(z)=z+sum (a_nz_n) from n=2 to +∞+sum(b_nz^n)from n=1 to +∞为定义在单位圆盘U上的调和映照,满足条件sum(np) from n=2 to +∞(|an|+|bn|)≤1-|b1|,证明当0<p≤1时,f(z)在圆盘|z|<r0=1/(21-p)内单叶;当1<p...设f(z)=h(z)+g(z)=z+sum (a_nz_n) from n=2 to +∞+sum(b_nz^n)from n=1 to +∞为定义在单位圆盘U上的调和映照,满足条件sum(np) from n=2 to +∞(|an|+|bn|)≤1-|b1|,证明当0<p≤1时,f(z)在圆盘|z|<r0=1/(21-p)内单叶;当1<p≤2时,(z)在圆盘|z|<R=1/(22-p)内为凸像函数.所得结果推广了M.Jahangiri等和M.ztürk等的结论.展开更多
Suppose that g(z)=z+a_2z^2+…is analytic for |z|<1, The condition Re{z g(z)/g(z)}>0 for |z|<1 (1) is necessary and sufficient for g(z) to be univalent and starlike for |z|<1. The condition Re{z g″(z)/g′(...Suppose that g(z)=z+a_2z^2+…is analytic for |z|<1, The condition Re{z g(z)/g(z)}>0 for |z|<1 (1) is necessary and sufficient for g(z) to be univalent and starlike for |z|<1. The condition Re{z g″(z)/g′(z)+1}>0 for |z|<1 (2) is necessary and sufficient for g(z) to be univalent and convex for |z|<1 In this paper we determine the radius Univalence (and starlikeness) of f(z) assodated with each of the cases (ⅰ) g(z) satisfies Re {g(z)/z}>0 for |z|<1 (ⅱ) g(z) satisfies eq .(1), (ⅲ) g(z) satisfies eq. (2). Definition. Let N_αdenote the class of functions f(z) which are regular in|z|<1 and are normalized by f(0)=1 and map |z|<1 into region D_α=展开更多
设f(z)=z+sum from p=2(a_pz^p)是单位圆|z|<1内的解析函数,记这种函数的全体为N.文[1]证明了:只要有|z|<1内单叶函数g(z)∈N(即g(z)∈S),使得Re{f(z)/g(z)}>0,则f(z)必在|z|<1/5内是单叶的.1980年吴卓人就g(z)属于S的一个...设f(z)=z+sum from p=2(a_pz^p)是单位圆|z|<1内的解析函数,记这种函数的全体为N.文[1]证明了:只要有|z|<1内单叶函数g(z)∈N(即g(z)∈S),使得Re{f(z)/g(z)}>0,则f(z)必在|z|<1/5内是单叶的.1980年吴卓人就g(z)属于S的一个子族,把上述结果加以完善.本文推广了吴卓人的这些结果.最后,还推广了MacGregor的另一个结果.展开更多
文摘设f(z)=h(z)+g(z)=z+sum (a_nz_n) from n=2 to +∞+sum(b_nz^n)from n=1 to +∞为定义在单位圆盘U上的调和映照,满足条件sum(np) from n=2 to +∞(|an|+|bn|)≤1-|b1|,证明当0<p≤1时,f(z)在圆盘|z|<r0=1/(21-p)内单叶;当1<p≤2时,(z)在圆盘|z|<R=1/(22-p)内为凸像函数.所得结果推广了M.Jahangiri等和M.ztürk等的结论.
文摘Suppose that g(z)=z+a_2z^2+…is analytic for |z|<1, The condition Re{z g(z)/g(z)}>0 for |z|<1 (1) is necessary and sufficient for g(z) to be univalent and starlike for |z|<1. The condition Re{z g″(z)/g′(z)+1}>0 for |z|<1 (2) is necessary and sufficient for g(z) to be univalent and convex for |z|<1 In this paper we determine the radius Univalence (and starlikeness) of f(z) assodated with each of the cases (ⅰ) g(z) satisfies Re {g(z)/z}>0 for |z|<1 (ⅱ) g(z) satisfies eq .(1), (ⅲ) g(z) satisfies eq. (2). Definition. Let N_αdenote the class of functions f(z) which are regular in|z|<1 and are normalized by f(0)=1 and map |z|<1 into region D_α=
文摘设f(z)=z+sum from p=2(a_pz^p)是单位圆|z|<1内的解析函数,记这种函数的全体为N.文[1]证明了:只要有|z|<1内单叶函数g(z)∈N(即g(z)∈S),使得Re{f(z)/g(z)}>0,则f(z)必在|z|<1/5内是单叶的.1980年吴卓人就g(z)属于S的一个子族,把上述结果加以完善.本文推广了吴卓人的这些结果.最后,还推广了MacGregor的另一个结果.
