2020屆一輪復(fù)習(xí)人教版數(shù)系的擴(kuò)充與復(fù)數(shù)的引入課件(38張)

1、第十四章 數(shù)系的擴(kuò)充與復(fù)數(shù)的引入第十四章 數(shù)系的擴(kuò)充與復(fù)數(shù)的引入20102019年高考全國(guó)卷考情一覽表 20102019年高考全國(guó)卷考情一覽表 2020屆一輪復(fù)習(xí)人教版-數(shù)系的擴(kuò)充與復(fù)數(shù)的引入-課件(38張)2020屆一輪復(fù)習(xí)人教版-數(shù)系的擴(kuò)充與復(fù)數(shù)的引入-課件(38張)2020屆一輪復(fù)習(xí)人教版-數(shù)系的擴(kuò)充與復(fù)數(shù)的引入-課件(38張)考點(diǎn)127考點(diǎn)128考點(diǎn)129考點(diǎn)127復(fù)數(shù)的四則運(yùn)算 2.(2019全國(guó)3,理2文2,5分,難度)若z(1+i)=2i,則z=(D)A.-1-iB.-1+iC.1-iD.1+i考點(diǎn)127考點(diǎn)128考點(diǎn)129考點(diǎn)127復(fù)數(shù)的四則運(yùn)算 2考點(diǎn)127考點(diǎn)128考點(diǎn)12

2、9考點(diǎn)127考點(diǎn)128考點(diǎn)129考點(diǎn)127考點(diǎn)128考點(diǎn)1296.(2018全國(guó)2,文1,5分,難度)i(2+3i)=(D)A.3-2iB.3+2iC.-3-2iD.-3+2i解析i(2+3i)=2i+3i2=-3+2i.7.(2018全國(guó)3,理2文2,5分,難度)(1+i)(2-i)=(D)A.-3-iB.-3+iC.3-iD.3+i解析(1+i)(2-i)=2+i-i2=3+i.考點(diǎn)127考點(diǎn)128考點(diǎn)1296.(2018全國(guó)2,文1,考點(diǎn)127考點(diǎn)128考點(diǎn)1298.(2018北京,理2文2,5分,難度)在復(fù)平面內(nèi),復(fù)數(shù) 的共軛復(fù)數(shù)對(duì)應(yīng)的點(diǎn)位于(D)A.第一象限B.第二象限C.第三象限D(zhuǎn)

3、.第四象限9.(2018浙江,4,4分,難度)復(fù)數(shù) (i為虛數(shù)單位)的共軛復(fù)數(shù)是(B)A.1+iB.1-iC.-1+iD.-1-i考點(diǎn)127考點(diǎn)128考點(diǎn)1298.(2018北京,理2文2考點(diǎn)127考點(diǎn)128考點(diǎn)12910.(2017全國(guó)1,理3,5分,難度)設(shè)有下面四個(gè)命題p2:若復(fù)數(shù)z滿足z2R,則zR; 其中的真命題為(B)A.p1,p3 B.p1,p4C.p2,p3 D.p2,p4 R.故p1正確;p2:因?yàn)閕2=-1R,而z=iR,故p2不正確;p3:若z1=1,z2=2,則z1z2=2,滿足z1z2R,而它們實(shí)部不相等,不是共軛復(fù)數(shù),故p3不正確;p4:實(shí)數(shù)的虛部為0,它的共軛復(fù)數(shù)

4、是它本身,也屬于實(shí)數(shù),故p4正確.考點(diǎn)127考點(diǎn)128考點(diǎn)12910.(2017全國(guó)1,理3考點(diǎn)127考點(diǎn)128考點(diǎn)12911.(2017全國(guó)2,理1,5分,難度) =(D)A.1+2i B.1-2iC.2+i D.2-i12.(2017全國(guó)2,文2,5分,難度)(1+i)(2+i)= (B)A.1-i B.1+3iC.3+i D.3+3i解析(1+i)(2+i)=2+3i+i2=1+3i,故選B.13.(2017山東,文2,5分,難度)已知i是虛數(shù)單位,若復(fù)數(shù)z滿足zi=1+i,則z2=(A)A.-2i B.2iC.-2 D.2考點(diǎn)127考點(diǎn)128考點(diǎn)12911.(2017全國(guó)2,理1考點(diǎn)1

5、27考點(diǎn)128考點(diǎn)129A.1B.-1C.iD.-I A.iB.1+iC.-iD.1-I 考點(diǎn)127考點(diǎn)128考點(diǎn)129A.1B.-1C.iD.考點(diǎn)127考點(diǎn)128考點(diǎn)12916.(2015全國(guó)2,理2,5分,難度)若a為實(shí)數(shù),且(2+ai)(a-2i)=-4i,則a=(B)A.-1B.0C.1D.217.(2015全國(guó)1,文3,5分,難度)已知復(fù)數(shù)z滿足(z-1)i=1+i,則z=(C)A.-2-iB.-2+iC.2-iD.2+i考點(diǎn)127考點(diǎn)128考點(diǎn)12916.(2015全國(guó)2,理2考點(diǎn)127考點(diǎn)128考點(diǎn)129a=(D)A.-4B.-3C.3D.4解析由題意,得2+ai=(3+i)(1

6、+i)=2+4i,則a=4.19.(2015安徽,文1,5分,難度)設(shè)i是虛數(shù)單位,則復(fù)數(shù)(1-i)(1+2i)=(C)A.3+3iB.-1+3iC.3+iD.-1+i解析由復(fù)數(shù)的乘法運(yùn)算法則,得(1-i)(1+2i)=1-i+2i-2i2=1+i+2=3+i,因此選C.考點(diǎn)127考點(diǎn)128考點(diǎn)129a=(D)19.(2015考點(diǎn)127考點(diǎn)128考點(diǎn)129z=(D)A.1+iB.1-iC.-1+iD.-1-iA.1+iB.1-iC.-1+iD.-1-I 考點(diǎn)127考點(diǎn)128考點(diǎn)129z=(D)A.1+iB.考點(diǎn)127考點(diǎn)128考點(diǎn)129A.1+2iB.-1+2iC.1-2iD.-1-2i 23

7、.(2013全國(guó)2,理2,5分,難度)設(shè)復(fù)數(shù)z滿足(1-i)z=2i,則z=(A)A.-1+iB.-1-iC.1+iD.1-i考點(diǎn)127考點(diǎn)128考點(diǎn)129A.1+2iB.-1+2i2考點(diǎn)127考點(diǎn)128考點(diǎn)129A.2-iB.1-2iC.-2+iD.-1+2i考點(diǎn)127考點(diǎn)128考點(diǎn)129A.2-iB.1-2i考點(diǎn)127考點(diǎn)128考點(diǎn)12927.(2018上海,5,4分,難度)已知復(fù)數(shù)z滿足(1+i)z=1-7i(i是虛數(shù)單位),則|z|=5.解析因?yàn)?1+i)z=1-7i,解得|z|=5. 28.(2017浙江,12,5分,難度)已知a,bR,(a+bi)2=3+4i(i是虛數(shù)單位),則a

8、2+b2=5,ab=2.解析由題意可得a2-b2+2abi=3+4i, 考點(diǎn)127考點(diǎn)128考點(diǎn)12927.(2018上海,5,4考點(diǎn)127考點(diǎn)128考點(diǎn)12929.(2016天津,理9,5分,難度)已知a,bR,i是虛數(shù)單位,若 考點(diǎn)127考點(diǎn)128考點(diǎn)12929.(2016天津,理9,考點(diǎn)127考點(diǎn)128考點(diǎn)129考點(diǎn)128復(fù)數(shù)的概念1.(2019全國(guó)2,文2,5分,難度)設(shè)z=i(2+i),則 =(D)A.1+2iB.-1+2iC.1-2iD.-1-2i2.(2017全國(guó)3,理2,5分,難度)設(shè)復(fù)數(shù)z滿足(1+i)z=2i,則|z|=(C)考點(diǎn)127考點(diǎn)128考點(diǎn)129考點(diǎn)128復(fù)數(shù)的概

9、念2.(2考點(diǎn)127考點(diǎn)128考點(diǎn)1293.(2017全國(guó)1,文3,5分,難度)下列各式的運(yùn)算結(jié)果為純虛數(shù)的是(C)A.i(1+i)2B.i2(1-i)C.(1+i)2 D.i(1+i)解析i(1+i)2=2i2=-2,i2(1-i)=-1+i,(1+i)2=2i,i(1+i)=-1+i,(1+i)2=2i為純虛數(shù),故選C.考點(diǎn)127考點(diǎn)128考點(diǎn)1293.(2017全國(guó)1,文3,考點(diǎn)127考點(diǎn)128考點(diǎn)1295.(2016全國(guó)1,理2,5分,難度)設(shè)(1+i)x=1+yi,其中x,y是實(shí)數(shù),則|x+yi|=(B)解析(定義、性質(zhì))因?yàn)?1+i)x=1+yi,x,yR,所以x=1,y=x=1.

10、所以|x+yi|=|1+i|= ,故選B.6.(2016全國(guó)1,文2,5分,難度)設(shè)(1+2i)(a+i)的實(shí)部與虛部相等,其中a為實(shí)數(shù),則a=(A)A.-3B.-2C.2D.3解析由已知(1+2i)(a+i)=a-2+(2a+1)i.(1+2i)(a+i)的實(shí)部與虛部相等,a-2=2a+1,解得a=-3,故選A.考點(diǎn)127考點(diǎn)128考點(diǎn)1295.(2016全國(guó)1,理2,考點(diǎn)127考點(diǎn)128考點(diǎn)1297.(2016全國(guó)2,文2,5分,難度)設(shè)復(fù)數(shù)z滿足z+i=3-i,則 =(C)A.-1+2iB.1-2iC.3+2iD.3-2i解析由z+i=3-i,得z=3-2i,所以 =3+2i,故選C.考

11、點(diǎn)127考點(diǎn)128考點(diǎn)1297.(2016全國(guó)2,文2,考點(diǎn)127考點(diǎn)128考點(diǎn)1299.(2016山東,理1,5分,難度)若復(fù)數(shù)z滿足2z+ =3-2i,其中i為虛數(shù)單位,則z=(B)A.1+2iB.1-2iC.-1+2iD.-1-2i考點(diǎn)127考點(diǎn)128考點(diǎn)1299.(2016山東,理1,5考點(diǎn)127考點(diǎn)128考點(diǎn)129(A)A.iB.-iC.1D.-1解析i607=i1514+3=i3=-i,i607的共軛復(fù)數(shù)為i.考點(diǎn)127考點(diǎn)128考點(diǎn)129(A)考點(diǎn)127考點(diǎn)128考點(diǎn)12913.(2013全國(guó)1,理2,5分,難度)若復(fù)數(shù)z滿足(3-4i)z=|4+3i|,則z的虛部為(D)考點(diǎn)1

12、27考點(diǎn)128考點(diǎn)12913.(2013全國(guó)1,理2考點(diǎn)127考點(diǎn)128考點(diǎn)129p1:|z|=2, p2:z2=2i,p3:z的共軛復(fù)數(shù)為1+i,p4:z的虛部為-1,其中的真命題為(C)A.p2,p3B.p1,p2C.p2,p4D.p3,p4A.2+iB.2-iC.-1+iD.-1-i考點(diǎn)127考點(diǎn)128考點(diǎn)129p1:|z|=2, 考點(diǎn)127考點(diǎn)128考點(diǎn)129考點(diǎn)127考點(diǎn)128考點(diǎn)129考點(diǎn)127考點(diǎn)128考點(diǎn)129考點(diǎn)127考點(diǎn)128考點(diǎn)129考點(diǎn)127考點(diǎn)128考點(diǎn)129命題點(diǎn)復(fù)數(shù)的運(yùn)算.解題思路先用復(fù)數(shù)的除法化簡(jiǎn),再求模.21.(2019江蘇,2,5分,難度)已知復(fù)數(shù)(a+2i

13、)(1+i)的實(shí)部為0,其中i為虛數(shù)單位,則實(shí)數(shù)a的值是2.解析(a+2i)(1+i)=a+ai+2i+2i2=a-2+(a+2)i,a-2=0,a=2.考點(diǎn)127考點(diǎn)128考點(diǎn)129命題點(diǎn)復(fù)數(shù)的運(yùn)算.21.(20考點(diǎn)127考點(diǎn)128考點(diǎn)12922.(2017江蘇,2,5分,難度)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 .24.(2016江蘇,2,5分,難度)復(fù)數(shù)z=(1+2i)(3-i),其中i為虛數(shù)單位,則z的實(shí)部是5.解析因?yàn)閦=(1+2i)(3-i)=5+5i,所以z的實(shí)部是5.考點(diǎn)127考點(diǎn)128考點(diǎn)12922.(2017江蘇,2,5考點(diǎn)127考點(diǎn)128考點(diǎn)

14、12925.(2015天津,理9,5分,難度)i是虛數(shù)單位,若復(fù)數(shù)(1-2i)(a+i)是純虛數(shù),則實(shí)數(shù)a的值為-2.解析(1-2i)(a+i)=a+2+(1-2a)i.(1-2i)(a+i)是純虛數(shù),a+2=0,且1-2a0,a=-2.26.(2015江蘇,3,5分,難度)設(shè)復(fù)數(shù)z滿足z2=3+4i(i是虛數(shù)單位),則z的模為 .27.(2015重慶,理11,5分,難度)設(shè)復(fù)數(shù)a+bi(a,bR)的模為 ,則(a+bi)(a-bi)=3.即a2+b2=3.于是(a+bi)(a-bi)=a2-(bi)2=a2+b2=3.考點(diǎn)127考點(diǎn)128考點(diǎn)12925.(2015天津,理9,考點(diǎn)127考點(diǎn)1

15、28考點(diǎn)129考點(diǎn)129復(fù)數(shù)的幾何意義1.(2019全國(guó)1,理2,5分,難度)設(shè)復(fù)數(shù)z滿足|z-i|=1,z在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)為(x,y),則(C)A.(x+1)2+y2=1B.(x-1)2+y2=1C.x2+(y-1)2=1D.x2+(y+1)2=1解析設(shè)z=x+yi(x,yR).因?yàn)閦-i=x+(y-1)i,則x2+(y-1)2=1.故選C. 考點(diǎn)127考點(diǎn)128考點(diǎn)129考點(diǎn)129復(fù)數(shù)的幾何意義解析考點(diǎn)127考點(diǎn)128考點(diǎn)1292.(2019全國(guó)2,理2,5分,難度)設(shè)z=-3+2i,則在復(fù)平面內(nèi) 對(duì)應(yīng)的點(diǎn)位于(C)A.第一象限B.第二象限C.第三象限D(zhuǎn).第四象限解析由z=-3+2i,得

16、 =-3-2i,則在復(fù)平面內(nèi) 對(duì)應(yīng)的點(diǎn)(-3,-2)位于第三象限,故選C.3.(2017全國(guó)3,文2,5分,難度)復(fù)平面內(nèi)表示復(fù)數(shù)z=i(-2+i)的點(diǎn)位于(C)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z=-1-2i,在復(fù)平面內(nèi)對(duì)應(yīng)點(diǎn)(-1,-2),則該點(diǎn)位于第三象限.故選C.考點(diǎn)127考點(diǎn)128考點(diǎn)1292.(2019全國(guó)2,理2,考點(diǎn)127考點(diǎn)128考點(diǎn)1294.(2017北京,理2,5分,難度)若復(fù)數(shù)(1-i)(a+i)在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)在第二象限,則實(shí)數(shù)a的取值范圍是(B)A.(-,1)B.(-,-1)C.(1,+)D.(-1,+)解析設(shè)z=(1-i)(a+i)=

17、(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)考點(diǎn)127考點(diǎn)128考點(diǎn)1294.(2017北京,理2,5考點(diǎn)127考點(diǎn)128考點(diǎn)1295.(2016全國(guó)2,理1,5分,難度)已知z=(m+3)+(m-1)i在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)在第四象限,則實(shí)數(shù)m的取值范圍是(A)A.(-3,1)B.(-1,3)C.(1,+)D.(-,-3)解析要使復(fù)數(shù)z在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)在第四象限,應(yīng)滿足考點(diǎn)127考點(diǎn)128考點(diǎn)1295.(2016全國(guó)2,理1,考點(diǎn)127考點(diǎn)128考點(diǎn)1296.(2015安徽,理1,5分,難度)設(shè)i是虛數(shù)單位,則復(fù)數(shù) 在復(fù)平面內(nèi)所對(duì)應(yīng)的點(diǎn)位于(B)A.第一象限B.第二象限C.第三象限D(zhuǎn).第四象限考點(diǎn)127考點(diǎn)128考點(diǎn)1296.(2015安徽,理1,5考點(diǎn)127考點(diǎn)128考點(diǎn)129