求积分计算f{|z|=pi}(z/(z+1))*(e^(2/(z+1)))dz
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求积分计算f{|z|=pi}(z/(z+1))*(e^(2/(z+1)))dz求积分计算f{|z|=pi}(z/(z+1))*(e^(2/(z+1)))dz求积分计算f{|z|=pi}(z/(z+1)
求积分计算f{|z|=pi}(z/(z+1))*(e^(2/(z+1)))dz
求积分计算f{|z|=pi}(z/(z+1))*(e^(2/(z+1)))dz
求积分计算f{|z|=pi}(z/(z+1))*(e^(2/(z+1)))dz
f(z)=z/(z+1)*e^[2/(z+1)]
设I=∫(|z|=π)f(z)dz
因为在区域|z|<=π内,f(z)只有一个奇点z0=-1,由留数定理,I等于2πi乘以f(z)在-1处的留数,又由于在该点的留数等于对应洛朗级数的1/(z+1)的系数a[-1],
而f(z)=[1-1/(z+1)]{1+2/(z+1)+(1/2!)[2/(z+1)]^2+...} (z≠-1)
因为只需求1/(z+1)的系数,观察可得:
a[-1]/(z+1)=1*2/(z+1)+[-1/(z+1)]*1=1/(z+1)
a[-1]=1
因此I=2πia[-1]=2πi
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