复数课堂巩固例题答案版.docx

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1、L下列运算结果是纯虚数的是(C )A(l + z)2Bj2(I-Z)C.(l + /)2Did + /)2为虚数单位，若复数(1+加)(i + 2)是纯虚数，则机=(D )AlR-IC.-2D23.若复数Z =义在复平面内对应的点在虚轴上, 1-/则实数”(B )A-2B.2G-IDO4. 3 的虚部为(B )1 + /A-zBTC.2/0.25.若复数z = 2i一旦是实数，贝J = ( D ) I-ZA-4B.2C.-2D.41-7；6.己知mH, i为虚数单位，若0,则加 m-i=(B )AlR-2C.-3D.-27.若Z = i(-2 + i),则z在复平面内对应的点位于(B )A.第

2、一象限B.第二象限C第三象限D第四象限8.若复数(l-Z)( + i)在复平面内对应的点位于第二象限，则实数的取值范围为(B )A(- 1)R(-,-l)C(l,+)D.(-l,+)9.若z = 4+3i,则 = ( D ) IZlAlR-I4 3.C.- + -/ 5 5八4 3.D.15 510.设复数(l + i)z = 2i,则 IZI = ( C )a 1R母22C.2D.211.若 zi = l + i,则 z2=( A )A-2ZB.2iC.-2D.212.若。为实数，且(2 + )(-2i) = Ti,则 =.(B )A-IRO13.若( + i)i = b-2i,则 + 6

3、= ( D )ClD.2AlR-IC.-2D.-32015FX C)22C 上 2D 土215.若亚=1-贝U Z的共规复数为（ ZA)Al-2zB.2-4iC.41-2411D.1 + 2Z16.复数 Z 满足(z-i)i = 2 +i,则 z =(B)A-1-zB-iC.-l + 3z).1-2;17.复数 Z 满足(z-2i)(2-i) = 5,则Z=(A)A2 + 3zB2-3iG3 + 2zD.3-2718.复数Z满足匕 =则z = ( A1 - z)AlB.C.3D.219.己知复数Z满足-Z = 2-4i,则z =(B)A3+ 4/R3-4/C.-3 + 4/D.-3-4z20.设复数Z满足z-l = l,则Z在复平面内对应的点Z(X,)，)满足的轨迹方程为(B )A(x + 1)2+=1B.(x-1)2 + =1 Cx2+(y-I)2 =1D.x2+(y+ 1)2 = 121.设复数Z满足z-l=IZT,则Z在复平面内对应的点Zay)满足的轨迹方程为(B )A,y = -xB.y = xC.(x-1) +(1) =1f.(x+l) (y +1) = 1