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設x=a+bi,則√(a^2+b^2)+a+bi=1+3i,[√(a^2+b^2)+a-1]+(b-3)i=0. 所以,b-3=0、√(a^2+b^2)+a-1=0,解得:b=3、a=-4. 所以,方程x+x=1+3i的解是x=-4+3i.
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