2024-2025學(xué)年高中數(shù)學(xué)第3章數(shù)系的擴(kuò)充與復(fù)數(shù)的引入3.2.1復(fù)數(shù)的加減與乘法運(yùn)算課時(shí)素養(yǎng)評(píng)價(jià)含解析蘇教版選修2-2

PAGE課時(shí)素養(yǎng)評(píng)價(jià)十一復(fù)數(shù)的加減與乘法運(yùn)算(25分鐘·60分)一、選擇題(每小題5分,共25分)1.已知z=11-20i,則1-2i-z等于 ()A.z-1 B.z+1C.-10+18i D.10-18i【解析】選C.1-2i-z=1-2i-(11-20i)=-10+18i.【補(bǔ)償訓(xùn)練】計(jì)算(3-5i)+(-4-i)-(3+4i)=____________.

【解析】(3-5i)+(-4-i)-(3+4i)=(3-4-3)+(-5-1-4)i=-4-10i.答案:-4-10i2.若復(fù)數(shù)z滿意z+(3-4i)=1,則z的虛部是 ()A.-2 B.4 C.3 【解析】選B.z=1-(3-4i)=-2+4i,所以z的虛部是4.3.復(fù)數(shù)(3i-1)·i的虛部是 ()A.-1 B.-3 C.3 D.1【解析】選A.(3i-1)·i=3i2-i=-3-i,所以虛部為-1.【補(bǔ)償訓(xùn)練】設(shè)f(z)=z-2i,z1=2+4i,z2=-1+2i,則f(z1-z2)=_______________.

【解析】由條件知z1-z2=3+2i,所以f(z1-z2)=f(3+2i)=3+2i-2i=3.答案:34.復(fù)數(shù)z=2+i,則zQUOTE= ()A.-3 B.QUOTE C.5 D.QUOTE【解析】選C.因?yàn)閦=2+i,所以QUOTE=2-i,所以zQUOTE=(2+i)(2-i)=4-i2=5.5.已知復(fù)數(shù)z=(1+i)(2-i),則QUOTE= ()A.3+i B.3-iC.-3+i D.-3-i【解析】選B.z=(1+i)(2-i)=3+i,所以QUOTE=3-i.二、填空題(每小題5分,共15分)6.設(shè)復(fù)數(shù)z=a+i(a∈R,i為虛數(shù)單位),若(1+i)·z為純虛數(shù),則a的值為____________.

【解析】(1+i)·z=(1+i)(a+i)=(a-1)+(1+a)i,由已知得a-1=0且1+a≠0,即a=1.答案:17.已知復(fù)數(shù)z1,z2滿意z2=z1-iQUOTE,且復(fù)數(shù)z2的實(shí)部是-1,則z2的虛部為____________.

【解析】設(shè)z1=a+bi(a,b∈R),則QUOTE=a-bi,所以z2=(a+bi)-i(a-bi)=(a-b)+(b-a)i,由題意知a-b=-1,所以b-a=1.答案:18.已知a,b∈R,且a-1+2ai=4+bi,則a=____________,b=____________.

【解析】由已知得QUOTE解得QUOTE答案:510三、解答題(每小題10分,共20分)9.若(-7i+5)-(9-8i)+(x+yi)=2,求x+y.【解析】(-7i+5)-(9-8i)+(x+yi)=(5-9+x)+(-7+8+y)i=(x-4)+(y+1)i.所以(x-4)+(y+1)i=2,所以QUOTE解得QUOTE所以x+y=5.10.設(shè)z是虛數(shù),ω=z+QUOTE是實(shí)數(shù),且-1<ω<2,(1)求|z|的值及z的實(shí)部的取值范圍;(2)設(shè)u=QUOTE,證明u為純虛數(shù).【解析】(1)因?yàn)閦是虛數(shù),所以可設(shè)z=x+yi,x,y∈R,且y≠0.所以ω=z+QUOTE=x+yi+QUOTE=x+yi+QUOTE=x+QUOTE+QUOTEi.因?yàn)棣厥菍?shí)數(shù)且y≠0,所以y-QUOTE=0,所以x2+y2=1,即|z|=1.此時(shí)ω=2x.因?yàn)?1<ω<2,所以-1<2x<2,從而有-QUOTE<x<1,即z的實(shí)部的取值范圍是QUOTE.(2)設(shè)z=x+yi,x,y∈R,且y≠0,由(1)知,x2+y2=1,-QUOTE<x<1,所以u(píng)=QUOTE=QUOTE=QUOTE=QUOTE=-QUOTEi.因?yàn)閤∈QUOTE,y≠0,所以QUOTE≠0,所以u(píng)為純虛數(shù).(20分鐘·40分)1.(5分)已知復(fù)數(shù)z=(QUOTE-i)i-(1+i)(1-i)(i為虛數(shù)單位),則復(fù)數(shù)z的共軛復(fù)數(shù)為 ()A.-1+QUOTEi B.-1-QUOTEiC.QUOTE-i D.-QUOTE-i【解析】選B.z=(QUOTE-i)i-(1+i)(1-i)=(1+QUOTEi)-2=-1+QUOTEi,所以QUOTE=-1-QUOTEi.2.(5分)已知x是純虛數(shù),y是實(shí)數(shù),且2x-1+i=y-(3-y)i,則x+y= ()A.-1 B.-QUOTEiC.-1-QUOTEi D.1+QUOTEi【解析】選C.設(shè)x=ai,a∈R且a≠0,則-1+(2a+1)i=y-(3-y)i,所以QUOTE解得QUOTE所以x=-QUOTEi,x+y=-1-QUOTEi.3.(5分)若復(fù)數(shù)z1=3+4i,z2=a+i,且z1·QUOTE是實(shí)數(shù)(其中QUOTE為z2的共軛復(fù)數(shù)),則實(shí)數(shù)a=_______________.

【解析】z1·QUOTE=(3+4i)(a-i)=3a+4+(4a-3)i,由于z1·QUOTE是實(shí)數(shù),所以4a-3=0,即a=QUOTE.答案:QUOTE4.(5分)已知復(fù)數(shù)z1=QUOTE-2mi,z2=-m+m2i,若z1+z2>0,則實(shí)數(shù)m=_______________.

【解析】z1+z2=(QUOTE-2mi)+(-m+m2i)=(QUOTE-m)+(m2-2m)i.因?yàn)閦1+z2>0,所以z1+z2為實(shí)數(shù)且大于0,所以QUOTE解得m=2.答案:2【誤區(qū)警示】本題易因?qū)1+z2>0的含義理解不清而導(dǎo)致無法求解.5.(10分)已知復(fù)數(shù)z=1+i,QUOTE是z的共軛復(fù)數(shù),實(shí)數(shù)a,b滿意az+2bQUOTE=(a+2z)2,求a,b的值.【解析】由題意知QUOTE=1-i,所以az+2bQUOTE=a(1+i)+2b(1-i)=(a+2b)+(a-2b)i,(a+2z)2=(a+2)2-4+4(a+2)i=a2+4a+4(a+2)i,于是QUOTE解得QUOTE或QUOTE6.(10分)設(shè)i為虛數(shù)單位,復(fù)數(shù)z和ω滿意zω+2iz-2iω+1=0,QUOTE-z=2i,求z和ω的值.【解析】因?yàn)镼UOTE-z=2i,所以z=QUOTE-2i.代入zω+2iz-2iω+1=0,得(QUOTE-2i)(ω+2i)-2iω+1=0,所以ωQUOTE