PAGEPAGE1考點(diǎn)規(guī)范練30復(fù)數(shù)一、基礎(chǔ)鞏固1.設(shè)i為虛數(shù)單位,則復(fù)數(shù)(1+i)2=()A.0 B.2 C.2i D.2+2i2.已知復(fù)數(shù)z=2-i,則z·z的值為()A.5 B.5 C.3 D.33.若復(fù)數(shù)(1-i)(a+i)在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)在其次象限,則實(shí)數(shù)a的取值范圍是()A.(-∞,1) B.(-∞,-1)C.(1,+∞) D.(-1,+∞)4.若復(fù)數(shù)z滿意2z+z=3-2i,其中i為虛數(shù)單位,則z=()A.1+2i B.1-2i C.-1+2i D.-1-2i5.若復(fù)數(shù)z=1+i,z為z的共軛復(fù)數(shù),則下列結(jié)論正確的是()A.z=-1-i B.z=-1+iC.|z|=2 D.|z|=26.已知復(fù)數(shù)z滿意iz=i+z,則z=()A.-12+B.-12C.12-D.127.設(shè)復(fù)數(shù)z1,z2在復(fù)平面內(nèi)的對(duì)應(yīng)點(diǎn)關(guān)于實(shí)軸對(duì)稱,z1=1+i,則z1z2=()A.2 B.-2 C.1+i D.1-i8.設(shè)z=1+i,則2z+z2等于(A.1+i B.-1+i C.-i D.-1-i9.已知復(fù)數(shù)z1=2+2i,z2=1-3i(i為虛數(shù)單位),則復(fù)數(shù)z12zA.第一象限 B.其次象限C.第三象限 D.第四象限10.已知復(fù)數(shù)z=(1+i)(1+2i),其中i是虛數(shù)單位,則z的模是.

11.已知i是虛數(shù)單位,則1-i(1+i12.已知a∈R,i為虛數(shù)單位,若a-i2+i為實(shí)數(shù),則a的值為二、實(shí)力提升13.若z=1+2i,則4izz-1A.1 B.-1C.i D.-i14.設(shè)復(fù)數(shù)z1=-1+3i,z2=1+i,則z1+z2A.-1-i B.1+i C.1-i D.-1+i15.已知a,b∈R,(a+bi)2=3+4i(i是虛數(shù)單位),則a2+b2=,ab=.

16.已知復(fù)數(shù)z=3+i(1-3i)2,z17.已知復(fù)數(shù)z1,z2滿意z1=m+(4-m2)i,z2=2cosθ+(λ+3sinθ)i(m,λ,θ∈R),并且z1=z2,則λ的取值范圍是.

18.已知復(fù)數(shù)z1=-1+2i,z2=1-i,z3=3-4i,它們?cè)趶?fù)平面上對(duì)應(yīng)的點(diǎn)分別為A,B,C,若OC=λOA+μOB(λ,μ∈R),則λ+μ的值是.

三、高考預(yù)料19.若z是z的共軛復(fù)數(shù),且滿意z(1-i)2=4+2i,則z=()A.-1+2i B.-1-2i C.1+2i D.1-2i

考點(diǎn)規(guī)范練30復(fù)數(shù)1.C解析由題意,(1+i)2=1+2i+i2=2i,故選C.2.A解析z·z=(2-i)·(2+i)=22-i2=4-(-1)=5,故選A.3.B解析設(shè)z=(1-i)(a+i)=(a+1)+(1-a)i,因?yàn)閺?fù)數(shù)z在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)(a+1,1-a)在其次象限,所以a+1<0,1-a4.B解析設(shè)z=a+bi(a,b∈R),則2z+z=3a+bi=3-2i,故a=1,b=-2,則z=1-2i,選B.5.D解析z=1-i,|z|=1+1=26.C解析∵iz=i+z,∴(1-i)z=-i,即z=-i1-i7.A解析由題意可知z2=1-i,故z1z2=(1+i)·(1-i)=2.故選A.8.A解析2z+z2=21+i+(1+i)2=2(1-i)(1+i)(1-i)+2i=9.B解析∵z1=2+2i,z2=1-3i,∴z12z2=(2+2i)∴復(fù)數(shù)z12z2在復(fù)平面內(nèi)所對(duì)應(yīng)的點(diǎn)的坐標(biāo)為-12510.10解析由已知得z=(1+i)(1+2i)=-1+3i,故|z|=(-1)211.-12-12i解析112.-2解析∵a-∴-a+25=0,即a=-13.C解析由題意知z=1-2i,則4izz-1=14.C解析∵z1=-1+3i,z2=1+i,∴z=2i=2i(-1-i)2=15.52解析由題意可得a2-b2+2abi=3+4i,則a2-b2=3,ab=2,解得a216.14解析∵z=3+i(故z=-34∴z·z=31617.-916,7解析由復(fù)數(shù)相等的充要條件可得m=2cosθ,4-m2=λ+3sinθ,化簡(jiǎn),得4-4cos2θ=λ+3sinθ,由此可得λ=-4cos2θ-3sinθ+4=-4(1-sin2θ)因?yàn)閟inθ∈[-1,1],所以4sin2θ-3sinθ∈-918.1解析由條件得OC=(3,-4),OA=(-1,2),OB=(1,-1),依據(jù)OC=λOA+μOB得(3,-4)=λ(-1