Question

复变函数的问题要详细的A,B,C,D解释

Answer 1

设z=x+yi,则Rez=x,Rez²=x²-y²,|z|=√(x²+y²)

看在原点是否连续,只需考虑z->0时,f(z)极限是否为0

A,f(z)=x/(1+√(x²+y²)),z->0,则x->0,∴f(z)->0 ∴连续

B,f(z)=x²/√(x²+y²),∵-1≤x/√(x²+y²)≤1,∴x->0时有x·x/√(x²+y²)->0 ∴连续

C,f(z)=(x²-y²)/(x²+y²),当z沿着(x,kx)->0时,f(z)->(1-k²)/(1+k²)≠0 ∴不连续

D,f(z)=(x²-y²)²/(x²+y²),∵-1≤(x²-y²)/(x²+y²)≤1,而z->0时有x²-y²->0

∴(x²-y²)·(x²-y²)/(x²+y²)->0 ∴连续

∴选 C