10.3复数的三角形式及其运算课后篇巩固提升基础巩固1.12(cos30°+isin30°)×2(cos60°+isin60°)×3(cos45°+isin45°)=()A.3√22+3√22iB.3√22−3√22iC.-3√22+3√22iD.-3√22−3√22i解析12(cos30°+isin30°)×2(cos60°+isin60°)×3(cos45°+isin45°)=12×2×3[cos(30°+60°+45°)+isin(30°+60°+45°)]=3(cos135°+isin135°)=3(-√22+√22i)=-3√22+3√22i.故选C.答案C2.(cosπ2+isinπ2)×3(cosπ6+isinπ6)=()A.32+3√32iB.32−3√32iC.-32+3√32iD.-32−3√32i解析(cosπ2+isinπ2)×3(cosπ6+isinπ6)=3[cos(π2+π6)+isin(π2+π6)]1=3(cos2π3+isin2π3)=-32+3√32i.故选C.答案C3.4(cosπ+isinπ)÷[2(cosπ3+isinπ3)]=()A.1+√3iB.1-√3iC.-1+√3iD.-1-√3i解析4(cosπ+isinπ)÷[2(cosπ3+isinπ3)]=2[cos(π-π3)+isin(π-π3)]=2(cos2π3+isin2π3)=-1+√3i.故选C.答案C4.2÷[2(cos60°+isin60°)]=()A.12+√32iB.12−√32iC.√32+12iD.√32−12i解析2÷2[(cos60°+isin60°)]=2(cos0°+isin0°)÷[2(cos60°+isin60°)]=cos(0°-60°)+isin(0°-60°)=cos(-60°)+isin(-60°)=12−√32i.2故选B.答案B5.9(cos3π+isin3π)÷[3(cos2π+isin2π)]=()A.3B.-3C.√3iD.-√3i解析9(cos3π+isin3π)÷[3(cos2π+isin2π)]=3[cos(3π-2π)+isin(3π-2π)]=3(cosπ+isinπ)=-3.故选B.答案B6.复数z=(sin25°+icos25°)3的三角形式是()A.cos195°+isin195°B.sin75°+icos75°C.cos15°+isin15°D.cos75°+isin75°解析z=(sin25°+icos25°)3=(cos65°+isin65°)3=cos195°+isin195°.故选A.答案A7.复数z=(cos40°+isin40°)6的结果是()A.12+√32iB.12−√32iC.-12+√32iD.-12−√32i解析z=(cos40°+isin40°)6=cos240°+isin240°3=-12−√32i.故选D.答案D8.2(cos15°+isin15°)×5(√32+12i)=.解析2(cos15°+isin15°)×5(√32+12i)=2(cos15°+isin15°)×5(cos30°+isin30°)=10[cos(15°+30°)+isin(15°+30°)]=10(cos45°+isin45°)=10(√22+√22i)=5√2+5√2i.答案5√2+5√2i9.12(cos60°-isin240°)×6(cos30°-isin210°)=.解析12(cos60°-isin240°)×6(cos30°-isin210°)=12(cos60°+isin60°)×6(cos30°+isin30°)=3[cos(60°+30°)+isin(60°+30°)]=3(cos90°+isin90°)=3i.答案3i10.2(cos210°+isin210°)×5(-sin30°+isin60°)=.解析2(cos210°+isin210°)×5(-sin30°+isin60°)=10(cos210°+isin210°)×(cos120°+isin120°)=10[cos(210°+120°)+isin(210°+120°)]=10(cos330°+isin330°)4=10(√32-12i)=5√3-5i.答案5√3-5i11.在复平面内,把与复数-2+2i对应的向量绕原点O按逆时针方向旋转75°,求与所得向量对应的复数.解所得向量对应的复数为(-2+2i)×(cos75°+isin75°)=2√2(cos135°+isin135°)×(cos75°+isin75°)=2√2[cos(135°+75°)+isin(135°+75°)]=2√2(cos210°+isin210°)=2√2(-√32-12i)=-√6−√2i.能力提升1.复数2+i和-3-i的辐角主值分别是α,β,则tan(α+β)等于()A.√3B.-√33C.-1D.1解析复数2+i和-3-i的辐角主值分别是α,β,所以tanα=12,tanβ=13,所以tan(α+β)=tanα+tanβ1-tanαtanβ=1.故选D.答案D2.复数-i的一个立方根是i,它的另外两个立方根是()A.√32±12iB.-√32±12i5C.±√32+12iD.±√32−12i解析-i=cos3π2+isin3π2∴-i的立方根为cos3π2+2kπ3+isin3π2+2kπ3(其中,k=0,1,2).当k=0时,得cosπ2+isinπ2=i.当k=1时,得cos7π6+isin7π6=-√32−12i.当k=2时,得cos11π6+isin11π6=√32−12i.故选D.答案D3.把复数z1与z2对应的向量⃗OA,⃗OB分别按逆时针方向旋转π4和5π3后,重合于向量⃗OM且模相等,已知z2=-1-√3i,则复数z1的代数式和它的辐角主值分别是()A.-√2+√2i,3π4B.-√2−√2i,3π4C.-√2+√2i,π4D.-√2−√2i,π4解析由复数乘法的几何意义得,z1(cosπ4+isinπ4)=z2(cos5π3+isin5π3).6又z2=-1-√3i=2(cos4π3+isin4π3),∴z1=2(cos4π3+isin4π3)(cos5π3+isin5π3)cosπ4+isinπ4=2[cos(3π-π4)+isin(3π-π4)]=-√2+√2i,z1的辐角主值为3π4.故选A.答案A4.在复平面内,复数z=a+bi(a∈R,b∈...
