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比如Log[z]在$\Bbb C\setminus(-∞,0]$解析
In[]:= FunctionAnalytic[{Log[z],!(Re[z]<=0&&Im[z]==0)},z]
Out[]= True
因此Log[z^2-z]在$z^2-z\nleq0$解析
In[]:= Reduce[z^2-z<=0,z,Complexes]
Out[]= (0<=Re[z]<1/2&&Im[z]==0)||Re[z]==1/2||(1/2<Re[z]<=1&&Im[z]==0)
显然$z^2-z\leq0$等价于$z\in[0,1]\cup\{z:\operatorname{Re}z=\frac12\}$
因此Log[z^2-z]在$\Bbb C\setminus[0,1]\setminus\{z:\operatorname{Re}z=\frac12\}$解析
In[]:= FunctionAnalytic[{Log[z^2-z], !(0 <= z <= 1 || Re[z] == 1/2)},z]
Out[]= True
In[]:= FunctionSingularities[Log[z^2-z],z,Complexes]
Out[]= -z+z^2==0||(Im[-z+z^2]==0&&Re[-z+z^2]<=0)
但是Log[z]+Log[z-1]在这个域不解析
In[]:= FunctionAnalytic[{Log[z]+Log[z-1], !(0 <= z <= 1 || Re[z] == 1/2)},z]
Out[]= False
Log[z]+Log[z-1]在$z\nleq0\land z-1\nleq0$解析, 即$\Bbb C\setminus(-∞,1]$解析
In[]:= FunctionAnalytic[{Log[z]+Log[z-1],!(z<=1)},z]
Out[]= True |
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