2020新教材高中数学 第十章 复数 10.3 复数的三角形式及其运算练习 新人教B版必修第四册

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.(cos30°+isin30°)×(cos60°+isin60°)×3(cos45°+isin45°)=()A.3√3√iB.3√3√iC.-3√3√iD.-3√3√i解析(cos30°+isin30°)×(cos60°+isin60°)×3(cos45°+isin45°)=×2×3[cos(30°+60°+45°)+isin(30°+60°+45°)]=3(cos35°+isin35°)=3(-√√i)=-3√3√i.故选C.答案C2.(cosisin)×3(cos6isin6)=()A.33√3iB.33√3iC.-33√3iD.-33√3i解析(cosisin)×3(cos6isin6)=3[cos(6)isin(6)]=3(cos3isin3)=-33√3i.故选C.答案C3.4(cos+isin)÷[(cos3isin3)]=()A.1+√3iB.1-√3iC.-1+√3iD.-1-√3i解析4(cos+isin)÷[(cos3isin3)]=2[cos(-3)isin(-3)]=2(cos3isin3)=-1+√3i.故选C.答案C4.2÷[(cos60°+isin60°)]=()A.√3iB.√3iC.√3iD.√3i解析2÷[(cos60°+isin60°)]=(cos0°+isin0°)÷[(cos60°+isin60°)]=cos(0°-60°)+isin(0°-60°)=cos(-60°)+isin(-60°)=√3i.故选B.答案B5.9(cos3+isin3)÷[3(cos2+isin2)]=()A.3B.-3C.√3iD.-√3i解析9(cos3+isin3)÷[3(cos2+isin2)]=3[cos(3-2)+isin(3-2)]=3(cos+isin)=-3.故选B.答案B6.复数z=(sin5°+icos5°)3的三角形式是()A.cos95°+isin95°B.sin75°+icos75°C.cos5°+isin5°D.cos75°+isin75°解析z=(sin5°+icos5°)3=(cos65°+isin65°)3=cos95°+isin95°.故选A.答案A7.复数z=(cos40°+isin40°)6的结果是()A.√3iB.√3iC.-√3iD.-√3i解析z=(cos40°+isin40°)6=cos40°+isin40°=-√3i.故选D.答案D8.(cos5°+isin5°)×5(√3i)=.解析(cos5°+isin5°)×5(√3i)=(cos5°+isin5°)×5(cos30°+isin30°)=0[cos(5°+30°)+isin(5°+30°)]=0(cos45°+isin45°)=10(√√i)=5√+5√i.答案5√+5√i9.(cos60°-isin40°)×6(cos30°-isin0°)=.解析(cos60°-isin40°)×6(cos30°-isin0°)=(cos60°+isin60°)×6(cos30°+isin30°)=3[cos(60°+30°)+isin(60°+30°)]=3(cos90°+isin90°)=3i.答案3i10.(cos0°+isin0°)×5(-sin30°+isin60°)=.解析(cos0°+isin0°)×5(-sin30°+isin60°)=0(cos0°+isin0°)×(cos0°+isin0°)=0[cos(0°+0°)+isin(0°+0°)]=0(cos330°+isin330°)=10(√3-i)=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√(cos35°+isin35°)×(cos75°+isin75°)=2√[cos(35°+75°)+isin(35°+75°)]=2√(cos0°+isin0°)=2√(-√3-i)=-√6√i.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=,tanβ=3,所以tan(α+β)=nn-nn=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.√3iB.-√3iC.±√3iD.±√3i解析-i=cos3+isin3∴-i的立方根为cos33+isin33(其中,k=0,1,2).当k=0时,得cos+isin=i.当k=1时,得cos76+isin76=-√3i.当k=2时,得cos6+isin6√3i.故选D.答案D3.把复数z1与z2对应的向量⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗分别按逆时针方向旋转4和53后,重合于向量⃗⃗⃗⃗⃗⃗⃗⃗⃗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.-√√i,34B.-√√i,34C.-√√i,4D.-√√i,4解析由复数乘法的几何意义得,z1(cos4isin4)=z2(cos53isin53).又z2=-1-√3i=2(cos43isin43),∴z1=(cos43isin43)(cos53isin53)cos4isin4=2[cos(3-4)isin(3-4)]=-√√i,z1的辐角主值为34.故选A.答案A4.在复平面内,复数z=a+bi(a∈R,b∈R)对应向量⃗⃗⃗⃗⃗⃗⃗⃗⃗(O为坐标原点),设|⃗⃗⃗⃗⃗⃗⃗⃗⃗|=r,以射线Ox为始边,OZ为终边旋转的角为θ,则z=r(cosθ+isinθ),法国数学家棣莫弗发现棣莫弗定理:z1=r1(cosθ1+isinθ1),z2=r2(cosθ2+isinθ2),则z1z2=r1r2[cos(θ1+θ2)+isin(θ1+θ2)],由棣莫弗定理导出了复数乘方公式:zn=[r(cosθ+isinθ)]n=rn(cosnθ+isinnθ),则(-1+√3i)10=()A.1024-104√3iB.-1024+1024√3iC.512-512√3iD.-512+512√3i解析根据复数乘方公式:zn=[r(cosθ+isinθ)]n=rn(cosnθ+isinnθ),得(-1+√3i)10=210[cos(03)isin(03)]=1024(cos03isin03)=1024(-√3i)=-512+512√3i.故选D.答案D5.设复数z=cos3+isin3,则--=()A.0B.√33iC.D.3解析--=--=--=-cos3-isin3cos3-isin3-isin3=sin3-isin3cos3-cos(3)isin(3)√3[cos()isin()]=cos0isin0sin3[cos(-6)isin(-6)]-√3[cos(6)isin(6)]=√3(cos6isin6-√3i)=√33i.故选B.答案B6.设(√3i)00=f(x)+ig(x)[f(x),g(x)均为实系数多项式],则f(x)的系数之和是()A.-√3B.√3C.-D.解析取x=1,则(√3i)00=f(x)+ig(x)⇒(cos6isin6)00=f(1)+ig(1)⇒(cos6isin6)4=f(1)+ig(1)⇒cos3+isin3=f(1)+ig(1)=-√3i.故选C.答案C7.6÷[3(cos35°+isin35°)]=.解析6÷3[(cos35°+isin35°)]=6(cos0°+isin0°)÷[3(cos35°+isin35°)]=[cos(0°-35°)+isin(0°-35°)]=4[cos(-35°)+isin(-35°)]=-2√-2√i.答案-2√-2√i8.已知复数z=cos3+isin3,则z3+=.解析根据题意,有z3+=1+z2=-z=√3i.答案√3i9.复数z=6(cos40°+isin40°)的四次方根分别是.解析z=6(cos40°+isin40°)的四次方根分别是√64(40°360°440°360°4)(k=0,1,2,3),当k=0时,(cos0°+isin0°);当k=1时,(cos00°+isin00°);当k=2时,(cos90°+isin90°);当k=3时,(cos0°+isin0°).答案(cos0°+isin0°),(cos00°+isin00°),(cos90°+isin90°),(cos0°+isin0°)10.设复数z1=√3+i,复数z2满足|z2|=2,已知z1的对应点在虚轴的负半轴上,且argz2∈(0,),求z2的代数形式.解因为z1=2(cos6isin6),设z2=2(cosα+isinα),α∈(0,),所以z1=8[cos(6)isin(6)].由题设知2α+6=2k+3(k∈Z),所以α=k+3(k∈Z).又α∈(0,),所以α=3.所以z2=2(cos3isin3)=-1+√3i.11.已知复数z=√3i,ω=√√i,复数,z2ω3在复平面上所对应的点分别为P,Q.求证:△OPQ是等腰直角三角形(其中O为原点).证明z=√3i=cos(-6)+isin(-6)ω=√√i=cos4+isin4,∴zω=cos(-64)+isin(-64)=cos+isin,∴=cos(-)+isin(-).又z2ω3=[cos(3)isin(3)](cos34isin34)=cos5+isin5,因此OP,OQ的夹角为5(-).∵OP⊥OQ,又∵|OP|=|z|=1,|OQ|=|z2|=1,∴|OP|=|OQ|,∴△OPQ为等腰直角三角形.