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设Q={f(z):f(z)=z-an+1zn+1-(∞∑k=n+2)akzk},这里an+1=c(n+2)/(n+1)(n+3),ak≥0,∞∑k=n+2k(k+2)/k+1ak≤1-c,0≤c≤1,n∈N,并且f(z)在单位圆盘△={z:| z |<1}内解析,得到函数族Q的极值点与支撑点.