復(fù)數(shù)的三角形式及其運算

第十章復(fù)數(shù)復(fù)數(shù)的三角形式及其運算課后篇鞏固提升基礎(chǔ)鞏固1.12(cos30°+isin30°)×2(cos60°+isin60°)×3(cos45°+isin45°)=(322322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-=-322故選C.答案C2.cosπ2+isinπ2×332+32+解析cosπ2+isin=3cos=3cos=-32+故選C.答案C(cosπ+isinπ)÷2cosπ3+isin+3i 3+3i 3解析4(cosπ+isinπ)÷2=2cos=2cos=-1+3i.故選C.答案C÷[2(cos60°+isin60°)]=()12+32+解析2÷2[(cos60°+isin60°)]=2(cos0°+isin0°)÷[2(cos60°+isin60°)]=cos(0°-60°)+isin(0°-60°)=cos(-60°)+isin(-60°)=12-故選B.答案B(cos3π+isin3π)÷[3(cos2π+isin2π)]=() 3 3解析9(cos3π+isin3π)÷[3(cos2π+isin2π)]=3[cos(3π-2π)+isin(3π-2π)]=3(cosπ+isinπ)=-3.故選B.答案B6.復(fù)數(shù)z=(sin25°+icos25°)3的三角形式是()195°+isin195°75°+icos75°15°+isin15°75°+isin75°解析z=(sin25°+icos25°)3=(cos65°+isin65°)3=cos195°+isin195°.故選A.答案A7.復(fù)數(shù)z=(cos40°+isin40°)6的結(jié)果是()12+12+解析z=(cos40°+isin40°)6=cos240°+isin240°=-12-故選D.答案D(cos15°+isin15°)×532+12解析2(cos15°+isin15°)×53=2(cos15°+isin15°)×5(cos30°+isin30°)=10[cos(15°+30°)+isin(15°+30°)]=10(cos45°+isin45°)=102=52+52i.答案52+52i9.12(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3i(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°)=103=53-5i.答案53-5i11.在復(fù)平面內(nèi),把與復(fù)數(shù)-2+2i對應(yīng)的向量繞原點O按逆時針方向旋轉(zhuǎn)75°,求與所得向量對應(yīng)的復(fù)數(shù).解所得向量對應(yīng)的復(fù)數(shù)為(-2+2i)×(cos75°+isin75°)=22(cos135°+isin135°)×(cos75°+isin75°)=22[cos(135°+75°)+isin(135°+75°)]=22(cos210°+isin210°)=22=-6-2能力提升1.復(fù)數(shù)2+i和-3-i的輻角主值分別是α,β,則tan(α+β)等于()A.3 3 解析復(fù)數(shù)2+i和-3-i的輻角主值分別是α,β,所以tanα=12,tanβ=1所以tan(α+β)=tanα+tanβ故選D.答案D2.復(fù)數(shù)-i的一個立方根是i,它的另外兩個立方根是()32±C.±32+12i D解析-i=cos3π2+∴-i的立方根為cos3π2+2kπ3+isin3當(dāng)k=0時,得cosπ2+isinπ2=當(dāng)k=1時,得cos7π6+isin7π6當(dāng)k=2時,得cos11π6+isin11故選D.答案D3.把復(fù)數(shù)z1與z2對應(yīng)的向量OA,OB分別按逆時針方向旋轉(zhuǎn)π4和5π3后,重合于向量OM且模相等,已知z2=-1-3i,2+22-22+22-2解析由復(fù)數(shù)乘法的幾何意義得,z1cos=z2cos5又z2=-1-3i=2cos4∴z1=2=2cos=-2+z1的輻角主值為3π故選A.答案A4.在復(fù)平面內(nèi),復(fù)數(shù)z=a+bi(a∈R,b∈R)對應(yīng)向量OZ(O為坐標原點),設(shè)|OZ|=r,以射線Ox為始邊,OZ為終邊旋轉(zhuǎn)的角為θ,則z=r(cosθ+isinθ),法國數(shù)學(xué)家棣莫弗發(fā)現(xiàn)棣莫弗定理:z1=r1(cosθ1+isinθ1),z2=r2(cosθ2+isinθ2),則z1z2=r1r2[cos(θ1+θ2)+isin(θ1+θ2)],由棣莫弗定理導(dǎo)出了復(fù)數(shù)乘方公式:zn=[r(cosθ+isinθ)]n=rn(cosnθ+isinnθ),則(-1+3i)10=()024-1043i 024+10243i3 +5123i解析根據(jù)復(fù)數(shù)乘方公式:zn=[r(cosθ+isinθ)]n=rn(cosnθ+isinnθ),得(-1+3i)10=210cos=1024cos=1024-=-512+5123i.故選D.答案D5.設(shè)復(fù)數(shù)z=cos23π+isin23π,則11- 3C.12 D.解析1=1=1=1=12sicos=cos0+isin02sin1=1=33i故選B.答案B6.設(shè)32+x2i2008=f(x)+ig(x)[f(x),g(x)均為實系數(shù)多項式],則32 B.32 12解析取x=1,則32+x2i2008=f(?cosπ6+isinπ62?cosπ6+isinπ64?cos2π3+isin2π3=f(1)+ig(1)故選C.答案C÷[3(cos135°+isin135°)]=.

解析6÷3[(cos135°+isin135°)]=6(cos0°+isin0°)÷[3(cos135°+isin135°)]=2[cos(0°-135°)+isin(0°-135°)]=4[cos(-135°)+isin(-135°)]=-22-22i.答案-22-22i8.已知復(fù)數(shù)z=cos2π3+isin2π3,則z3+z解析根據(jù)題意,有z3+z2z2+z+2=1答案129.復(fù)數(shù)z=16(cos40°+isin40°)的四次方根分別是.

解析z=16(cos40°+isin40°)的四次方根分別是4(k=0,1,2,3),當(dāng)k=0時,2(cos10°+isin10°);當(dāng)k=1時,2(cos100°+isin100°);當(dāng)k=2時,2(cos190°+isin190°);當(dāng)k=3時,2(cos280°+isin280°).答案2(cos10°+isin10°),2(cos100°+isin100°),2(cos190°+isin190°),2(cos280°+isin280°)10.設(shè)復(fù)數(shù)z1=3+i,復(fù)數(shù)z2滿足|z2|=2,已知z1·z22的對應(yīng)點在虛軸的負半軸上,且argz2∈(0,π),求z2解因為z1=2cosπ設(shè)z2=2(cosα+isinα),α∈(0,π),所以z1z22=8由題設(shè)知2α+π6=2kπ+3π2(k所以α=kπ+2π3(k∈Z又α∈(0,π),所以α=2π所以z2=2cos2π3+isin2π11.已知復(fù)數(shù)z=32-12i,ω=22+22i,復(fù)數(shù)zω,z2ω3