1) (1+2i)x+(3-10i)y=5-6i即x+3y=5,2x-10y=-6解得:x=2,y=12) x^2+xi+2-3i=y^2+yi+9-2i即 x^2+2=y^2+9,x-3=y-2解得:x=4,y=33) 2x^2-5x+3+(y^2+y-6)i=0即2x^2-5x+3=0,y^2+y-6=0解得:x=1或3/2,y=-3或24) x/(1-i)+y/(1-2i)=5/(1-3i)即(1+i)x/2+(1+2i)y/5=5(1+3i)/10x/2+y/5=5/10,x/2+2/5y=15/10解得:x=-1,y=51) 如果让实数a与纯虚数ai对应,那么实数集R与纯虚数集一一对应;
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