设复数z1=1+iz2=x+2i(x∈R)若z1z2为实数则x等于()A.-2B.-1C.1D.2
A.-2 B.-1 C.1 D.2
A
解析:z 1 z 2 =(1+i)(x+2i)=x-2+(x+2)i 若z 1 z 2 为实数 则虚部为0得x=-2.
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