學(xué)業(yè)分層測(cè)評(píng)第3章313

1、學(xué)業(yè)分層測(cè)評(píng)(建議用時(shí)：45分鐘)學(xué)業(yè)達(dá)標(biāo)一、選擇題1.在復(fù)平面內(nèi)，復(fù)數(shù)z=sin2+icos2對(duì)應(yīng)的點(diǎn)位于A. 第一象限B.第二象限C.第三象限D(zhuǎn).第四象限【解析】.Sin2>0,cos2<0,二復(fù)數(shù)z對(duì)應(yīng)的點(diǎn)(sin2,cos2廊第四象限.故選D.【答案】D2.已知復(fù)數(shù)z=(a22a)+(a2a2)i對(duì)應(yīng)的點(diǎn)在虛軸上，M()A.a2或a1B.a2,且a1C.a=0D.a=2或a=0【解析】由題意，得a2-2a=0,得a=0或a=2.故選D.【答案】D在復(fù)平面內(nèi),O為原點(diǎn)，向量OA對(duì)應(yīng)的復(fù)數(shù)為一1+2i,若點(diǎn)A關(guān)于直線y=x的對(duì)稱(chēng)點(diǎn)為點(diǎn)B,則向量OB對(duì)應(yīng)的復(fù)數(shù)為()A.-2-iB

2、.-2+iC.1+2iD.1+2i【解析】因?yàn)閺?fù)數(shù)一1+2i對(duì)應(yīng)的點(diǎn)為A(-1,2)，點(diǎn)A關(guān)于直線y=x的對(duì)稱(chēng)點(diǎn)為B(-2,1),所以O(shè)B對(duì)應(yīng)的復(fù)數(shù)為一2+i.【答案】B3. 已知復(fù)數(shù)z滿(mǎn)足|z|22Z|3=0,則復(fù)數(shù)z對(duì)應(yīng)點(diǎn)的軌跡是()A.1個(gè)圓B.線段C.2個(gè)點(diǎn)D.2個(gè)圓【解析】由題意知(|z|3)(|z|+1)=0,即|z|=3或|z|=1,|z|>0,.|z|=3，復(fù)數(shù)z對(duì)應(yīng)點(diǎn)的軌跡是1個(gè)圓.【答案】A實(shí)部為一2,虛部為1的復(fù)數(shù)所對(duì)應(yīng)的點(diǎn)位于復(fù)平面的()【導(dǎo)學(xué)號(hào)：05410066】A.第一象限B.第二象限C.第三象限D(zhuǎn).第四象限【解析】由題意可得復(fù)數(shù)z=2+i,故在復(fù)平面內(nèi)對(duì)應(yīng)的

3、點(diǎn)為(2,1),在第二象限，故選B.【答案】B二、填空題4. i為虛數(shù)單位，設(shè)復(fù)數(shù)z，z2在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)關(guān)于原點(diǎn)對(duì)稱(chēng)，若z=23i，貝Uz2=.【解析】復(fù)數(shù)z=2-3i對(duì)應(yīng)的點(diǎn)為(2,-3)，貝Uz2對(duì)應(yīng)的點(diǎn)為(一2,3),所以z2=2+3i.【答案】一2+3i5. 已知在zABC中，AB,AC對(duì)應(yīng)的復(fù)數(shù)分別為一1+2i,-2-3i,則BC對(duì)應(yīng)的復(fù)數(shù)為.【解析】因?yàn)锳fe,AC對(duì)應(yīng)的復(fù)數(shù)分別為一1+2i,-2-3i,所以AB=(-1,2),AC=(-2,3),乂BC=ACAB=(2,-3)-(-1,2)=(-1,5),所以BC對(duì)應(yīng)的復(fù)數(shù)為一15i.【答案】一15i6. 已知3-4i=x+y

4、i(x,yR)，貝U|15i|,|xyi|,|y+2i|的大小關(guān)系為【解析】由3-4i=x+yi(x,y£R),得x=3,y=4.而|1-5i|=寸+52=例，Xyi|=|3+4i|=V32+42=5,|y+2i|=|-4+2i|=/(42+22=伽，.20<5<.26,.|y+2i|<|x-yi|<|1-5i|.【答案】|y+2i|<|x-yi|<|15i|三、解答題7. 如果復(fù)數(shù)z=(m2+m1)+(4m28m+3)i(mR)對(duì)應(yīng)的點(diǎn)在第一象限，求實(shí)數(shù)m的取值范圍.【解】.復(fù)數(shù)z對(duì)應(yīng)的點(diǎn)在第一象限.m2+m1>0,24m8m+3>0

5、,153解得m<2或m)所以實(shí)數(shù)m的取值范圍為''-1-饞*3工、8,戶(hù)2+勺8. 已知x,yR，若x2+2x+(2y+x)i和3x-(y+1)i是共軸復(fù)數(shù)，求復(fù)數(shù)z=x+yi和z.【解】若兩個(gè)復(fù)數(shù)a+bi與c+di共鑰，貝Ua=c,且b=d.x+2x=3x,由此可得到關(guān)丁x,y的方程組S2y+x=y+1,x=0,解得y=ix=1,z=i,z=1,或4所以S或fy=0,z=iz=1.能力提升已知復(fù)數(shù)z對(duì)應(yīng)的向量為。成(。為坐標(biāo)原點(diǎn))，O芝與實(shí)軸正向的火角為120°，且復(fù)數(shù)z的模為2,則復(fù)數(shù)z為()A.1+船B.2C.(1,V3)D.1+V3i【解析】設(shè)復(fù)數(shù)z對(duì)應(yīng)

6、的點(diǎn)為(x,y),則x=|z|cos120=2X1,2尸T，y=|z|sin120=2X寫(xiě)=g復(fù)數(shù)z對(duì)應(yīng)的點(diǎn)為(一1,V3),n1+V3i.【答案】D1. 與x軸同方向的單位向量e,與y軸同方向的單位向量2,它們對(duì)應(yīng)的復(fù)數(shù)分別是()A. e1對(duì)應(yīng)實(shí)數(shù)1,e2對(duì)應(yīng)虛數(shù)iB. e1對(duì)應(yīng)虛數(shù)i,2對(duì)應(yīng)虛數(shù)iC. e1對(duì)應(yīng)實(shí)數(shù)1,e2對(duì)應(yīng)虛數(shù)一iD. e1對(duì)應(yīng)實(shí)數(shù)1或1,e2對(duì)應(yīng)虛數(shù)i或一i【解析】ei=(1,0),e2=(0,1).【答案】A2. 復(fù)數(shù)z=5-12i在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)到原點(diǎn)的距離為.【導(dǎo)學(xué)號(hào)：05410067【解析】復(fù)數(shù)z=5-12i在復(fù)平面內(nèi)對(duì)應(yīng)點(diǎn)Z(-5,12),所以點(diǎn)Z與原點(diǎn)O的距離為|OZ|=V(-5f+(-12f=13.【答案】134.已知。為坐標(biāo)原點(diǎn),OZi對(duì)應(yīng)的復(fù)數(shù)為一3+4i,OZ2對(duì)應(yīng)的復(fù)數(shù)為2a+i(aR).若OZi與OZ2共線，求a的值.【解】因?yàn)镺Zi對(duì)應(yīng)的復(fù)數(shù)為一3+4i,OZ2對(duì)應(yīng)的復(fù)數(shù)為2a+i,所以O(shè)Zi=(-3,4),&a