十七數(shù)系的擴(kuò)充和復(fù)數(shù)的概念(時(shí)間:45分鐘分值:90分)【基礎(chǔ)全面練】1.(5分)(2024·銅仁高一檢測(cè))復(fù)數(shù)z=-2i+i2,則復(fù)數(shù)z的虛部是()A.-2 B.2 C.-1 D.1【解析】選A.由題意可得:z=-2i+i2=-2i-1=-1-2i,所以復(fù)數(shù)z的虛部是-2.2.(5分)設(shè)C={復(fù)數(shù)},A={實(shí)數(shù)},B={純虛數(shù)},全集U=C,則下面結(jié)論正確的是()A.A∪B=C B.UA=BC.A∩(UB)=? D.B∪(UB)=C【解析】選D.由復(fù)數(shù)的分類可知D項(xiàng)正確.3.(5分)若復(fù)數(shù)(a+1)+(a2-1)i(a∈R)是實(shí)數(shù),則a等于()A.-1 B.1 C.±1 D.不存在【解析】選C.由題設(shè)a2-1=0,故a=±1.4.(5分)(2024·石家莊高一檢測(cè))已知i為虛數(shù)單位,a,b∈R,集合A=z|z=a+(2a-1)A.{2i} B.{1+3i} C.{3+5i} D.{2+4i}【解析】選C.由題得a+(2a-1)i=b-2+bi,所以a=b-22a-1=5.(5分)下列結(jié)論正確的是()A.z=1+i2是虛數(shù)B.z=3+i為純虛數(shù)C.z=3+i的實(shí)部大于虛部D.z=1-2i的虛部為-2i【解析】選C.z=1+i2=1-1=0,A不正確;z=3+i不是純虛數(shù),B不正確;z=3+i的實(shí)部為3,虛部為1,所以實(shí)部大于虛部,C正確;z=1-2i的虛部為-2,D不正確.【補(bǔ)償訓(xùn)練】若復(fù)數(shù)z=cosθ-45+sinθ-35i是純虛數(shù)(i為虛數(shù)單位),則tanθ-π4的值為()A.7 B.-1C.-7 D.-7或-1【解析】選C.因?yàn)閺?fù)數(shù)z=cosθ-45+sinθ-35i是純虛數(shù),所以cosθ-45=0,sinθ-35又cos2θ+sin2θ=1,所以cosθ=45,sinθ=-35,所以tanθ=-所以tanθ-π4=tanθ-11+tanθ6.(5分)(多選)對(duì)于復(fù)數(shù)a+bi(a,b∈R),下列說法正確的是()A.若a=0,則a+bi為純虛數(shù)B.若a+(b-1)i=3-2i,則a=3,b=-1C.若b=0,則a+bi為實(shí)數(shù)D.若b≠0,則a+bi是虛數(shù)【解析】選BCD.對(duì)于選項(xiàng)A,當(dāng)a=b=0時(shí),a+bi=0顯然是實(shí)數(shù),因此本選項(xiàng)說法不正確;對(duì)于選項(xiàng)B,a+(b-1)i=3-2i?a=3b-對(duì)于選項(xiàng)C,b=0,a+bi=a∈R,因此本選項(xiàng)說法正確;D選項(xiàng)說法正確.7.(5分)若a,b∈R,i是虛數(shù)單位,a+2024i=2-bi,則a2+bi等于__________.

【解析】因?yàn)閍+2024i=2-bi,所以a=2,-b=2024,即a=2,b=-2024,所以a2+bi=4-2024i.答案:4-2024i8.(5分)若z=(m2-1)+(m-1)i(m∈R)是純虛數(shù),則復(fù)數(shù)z的實(shí)部與虛部的和是______________.

【解析】因?yàn)閦是純虛數(shù),所以m2解得m=-1,從而復(fù)數(shù)z的實(shí)部與虛部分別是0和-2,其和是-2.答案:-29.(5分)(2024·廈門高一檢測(cè))已知復(fù)數(shù)z=m-3+(m2-9)i(i為虛數(shù)單位),若z≥0,則實(shí)數(shù)m的值為__________.

【解析】復(fù)數(shù)z=m-3+(m2-9)i(i為虛數(shù)單位),z≥0,則m-3≥0m2答案:310.(10分)設(shè)復(fù)數(shù)z=(a2+a-2)+(a2-7a+6)i,其中a∈R,當(dāng)a取何值時(shí),(1)z∈R;(2)z是純虛數(shù);(3)z是零.【解析】(1)若z∈R,則a2-7a+6=0,解得a=1或a=6.(2)若z是純虛數(shù),則a2+a-(3)若z是零,則a2+a-【補(bǔ)償訓(xùn)練】求實(shí)數(shù)m取什么數(shù)值時(shí),復(fù)數(shù)z=m2-m-2+(m2-1)i分別是:(1)實(shí)數(shù);(2)虛數(shù);(3)純虛數(shù).【解析】(1)當(dāng)m2-1=0,即m=1或m=-1時(shí),復(fù)數(shù)z是實(shí)數(shù);(2)當(dāng)m2-1≠0,即m≠1且m≠-1時(shí),復(fù)數(shù)z是虛數(shù);(3)當(dāng)m2即m=2時(shí),復(fù)數(shù)z是純虛數(shù).【綜合應(yīng)用練】11.(5分)已知復(fù)數(shù)z=a2+(2a+3)i(a∈R)的實(shí)部大于虛部,則實(shí)數(shù)a的取值范圍是()A.(-1,3)B.(-∞,-1)∪(3,+∞)C.(-3,1)D.(-∞,-3)∪(1,+∞)【解析】選B.由已知可得a2>2a+3,即a2-2a-3>0,解得a>3或a<-1,因此,實(shí)數(shù)a的取值范圍是(-∞,-1)∪(3,+∞).12.(5分)(多選)(2024·長春高一檢測(cè))已知復(fù)數(shù)z1=m2-1+(m+1)i,z2=cos2θ+isinθ,下列說法正確的是()A.若z1為純虛數(shù),則m=1B.若z2為實(shí)數(shù),則θ=kπ,k∈ZC.若z1=z2,則m=0或m=-4D.若z1≥0,則m的取值范圍是(-∞,-1]∪[1,+∞)【解析】選ABC.對(duì)于A,復(fù)數(shù)z1=m2-1+(m+1)i是純虛數(shù),則m2-1=0對(duì)于B,若z2=cos2θ+isinθ為實(shí)數(shù),則sinθ=0,則θ=kπ,k∈Z,B正確;對(duì)于C,若z1=z2,則m2-1=cos2θm+1=sinθ,則m2-1=1-2(m+1)2對(duì)于D,若z1≥0,則m2-1≥0,且m+1=0,則m=-1,D錯(cuò)誤.13.(5分)定義運(yùn)算abcd=ad-bc,如果(x+y)+(x+3)i=3x+2【解析】由題意得,(x+y)+(x+3)i=(3x+2y)+yi,所以x+y=3答案:-1,214.(10分)分別求滿足下列條件的實(shí)數(shù)x,y的值.(1)2x-1+(y+1)i=x-y+(-x-y)i;(2)x2-x-6x+1【解析】(1)因?yàn)閤,y∈R,2x-1+(y+1)i=x-y+(-x-y)i,則有2x解得x=3(2)因?yàn)閤∈R,x2-x-6x所以x2解得x=3.【補(bǔ)償訓(xùn)練】已知復(fù)數(shù)z1=m+(4-m2)i,m∈R,z2=2cosθ+(λ+3sinθ)i,λ∈R,θ∈R,且z1=z2,求λ的取值范圍.【解析】因?yàn)閦1=z2,所以m所以λ=4-4cos2θ-3sinθ=4sin2θ-3sinθ=4(sinθ-3因?yàn)?1≤sinθ≤1,所以λ∈-916,7.15.(10分)當(dāng)實(shí)數(shù)m為何值時(shí),復(fù)