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令f(z)=1+z+z^2+…+z^(k-1),f(z)=(1-z^k)/(1-z),|f(z)|=|1-z^k|/|1-z|,由於|z|<1,所以|z^k|=|z|^k<1,所以|f(z)|≥(1-|z^k|)/(1+|z|)>(1-|z|^k)/2>0,即|f(z)|>0,所以z=0一定不是多邊形的頂點….
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