f( i )=0则u(x,y)=u(i,y)=0且v(x,y)=v(i,y)=0
由u(x,y)=(x^2)-(y^2)+1得u(i,y)=i^2-y^2+1=0
∴y^2=i^2+1=0,y=0
则f(i)=0时,f(x)=u(i,0)+iv(i,0)
解析函数f(x)=u(x,y)+iv(x,y)的实部u(x,y)=(x^2)-(y^2)+1求满足条件f( i )=0的f(x)=u(x,y)+iv(x,y)
解析函数f(x)=u(x,y)+iv(x,y)的实部u(x,y)=(x^2)-(y^2)+1求满足条件f( i )=0的f(x)=u(x,y)+iv(x,y)
数学人气:648 ℃时间:2020-05-01 18:52:20
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