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設z=a+bi(a,b∈R),∵Z2+Z=0,∴(a+bi)2+a2+b2=0,∴a2-b2+a2+b2+2abi=0,∴a2−b2+a2+b2=02ab=0,解得,a=0或b=0,1,-1.則z=0或i或-i.故選C.
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