2024-2025學(xué)年新教材高中數(shù)學(xué)第12章復(fù)數(shù)12.1復(fù)數(shù)的概念課時(shí)素養(yǎng)評(píng)價(jià)含解析蘇教版必修第二冊(cè)

PAGE課時(shí)素養(yǎng)評(píng)價(jià)十八復(fù)數(shù)的概念(20分鐘35分)1.(2024·濰坊高一檢測)下列命題正確的是 ()A.(-i)2=-1B.-i2=-1C.若a>b,則a+i>b+iD.若z∈C,則z2>0【解析】選A.QUOTE=i2=-1,A正確;-i2=-QUOTE=1,B錯(cuò)誤;虛數(shù)無法比較大小,C錯(cuò)誤;若z=i,則z2=-1<0,D錯(cuò)誤.【補(bǔ)償訓(xùn)練】若a,b∈R,且a>b,那么 ()A.ai>bi B.a+i>b+iC.ai2>bi2 D.bi2>ai2【解析】選D.虛數(shù)不能比較大小,故A,B錯(cuò);因?yàn)閕2=-1,a>b,所以ai2<bi2,故C錯(cuò)D對(duì).2.假如z=m(m+1)+(m2-1)i為純虛數(shù),則實(shí)數(shù)m的值為 ()A.1 B.0C.-1 D.-1或1【解析】選B.由題意知QUOTE解得m=0.3.適合x-3i=(8x-y)i的實(shí)數(shù)x,y的值為 ()A.x=0且y=3 B.x=0且y=-3C.x=5且y=3 D.x=3且y=0【解析】選A.依題意得QUOTE解得QUOTE4.以QUOTEi-QUOTE的虛部為實(shí)部,以8i2+QUOTEi的實(shí)部為虛部的復(fù)數(shù)是________.

【解析】QUOTEi-QUOTE的虛部為QUOTE,8i2+QUOTEi=-8+QUOTEi的實(shí)部為-8.答案:QUOTE-8i5.滿意方程x2-2x-3+(9y2-6y+1)i=0的實(shí)數(shù)對(duì)(x,y)表示的點(diǎn)的個(gè)數(shù)為________.

【解析】由題意知QUOTE解得QUOTE或QUOTE所以實(shí)數(shù)對(duì)(x,y)表示的點(diǎn)有QUOTE,QUOTE,共有2個(gè).答案:26.已知復(fù)數(shù)z=(m2+3m+2)+(m2-m-6)i,則當(dāng)實(shí)數(shù)m為何值時(shí),復(fù)數(shù)z(1)是實(shí)數(shù);(2)是虛數(shù);(3)是純虛數(shù).【解析】(1)當(dāng)m2-m-6=0,即m=3或m=-2時(shí),z為實(shí)數(shù).(2)當(dāng)m2-m-6≠0,即m≠-2且m≠3時(shí),z是虛數(shù).(3)當(dāng)QUOTE即m=-1時(shí),z是純虛數(shù).(30分鐘60分)一、單選題(每小題5分,共20分)1.(2024·北京高一檢測)已知i是虛數(shù)單位,a,b∈R,“a=0”是“復(fù)數(shù)a+bi是純虛數(shù)”的 ()A.充分不必要條件B.必要不充分條件C.充分必要條件D.既不充分也不必要條件【解析】選B.復(fù)數(shù)a+bi是純虛數(shù),則QUOTE所以“a=0”是“復(fù)數(shù)a+bi是純虛數(shù)”的必要不充分條件.2.(2024·武漢高一檢測)已知a為實(shí)數(shù),若復(fù)數(shù)z=(a2-9)+(a+3)i為純虛數(shù),則復(fù)數(shù)z的虛部為 ()A.3 B.6i C.±3 D.6【解析】選D.因?yàn)閦=(a2-9)+(a+3)i為純虛數(shù),所以QUOTE解得a=3,所以z=6i,則復(fù)數(shù)z的虛部為6.3.(2024·三亞高二檢測)已知復(fù)數(shù)z=x+yiQUOTE,則 ()A.z2≥0B.z的虛部是yiC.若z=1+2i,則x=1,y=2D.z為實(shí)數(shù)時(shí),x+y=0【解析】選C.對(duì)于A選項(xiàng),取z=i,則z2=-1<0,A選項(xiàng)錯(cuò)誤;對(duì)于B選項(xiàng),復(fù)數(shù)z的虛部為y,B選項(xiàng)錯(cuò)誤;對(duì)于C選項(xiàng),若z=1+2i,則x=1,y=2,C選項(xiàng)正確;對(duì)于D選項(xiàng),z為實(shí)數(shù)時(shí),y=0,D選項(xiàng)錯(cuò)誤.4.若復(fù)數(shù)cosα+isinα和sinα+icosα相等,則α的值為 ()A.QUOTE B.QUOTE或QUOTEC.2kπ+QUOTE(k∈Z) D.kπ+QUOTE(k∈Z)【解析】選D.依據(jù)兩個(gè)復(fù)數(shù)相等的充要條件建立關(guān)系式得cosα=sinα,即tanα=1,所以α=kπ+QUOTE(k∈Z).二、多選題(每小題5分,共10分,全部選對(duì)得5分,選對(duì)但不全的得3分,有選錯(cuò)的得0分)5.對(duì)于復(fù)數(shù)z=a+bi(a,b∈R),下列結(jié)論錯(cuò)誤的是 ()A.若a=0,則a+bi為純虛數(shù)B.若a-bi=3+2i,則a=3,b=2C.若b=0,則a+bi為實(shí)數(shù)D.z為虛數(shù)時(shí),b≠0【解析】選AB.因?yàn)閦=a+bi(a,b∈R),當(dāng)a=0且b≠0時(shí)復(fù)數(shù)為純虛數(shù),故A錯(cuò)誤;當(dāng)b=0時(shí),復(fù)數(shù)為實(shí)數(shù),故C正確;z為虛數(shù)時(shí),b≠0,故D正確;對(duì)于B:a-bi=3+2i,則QUOTE即QUOTE故B錯(cuò)誤.6.(2024·揚(yáng)州高二檢測)已知i為虛數(shù)單位,下列說法正確的是 ()A.若x,y∈R,且x+yi=1+i,則x=y=1B.隨意兩個(gè)虛數(shù)都不能比較大小C.若復(fù)數(shù)z1,z2滿意QUOTE+QUOTE=0,則z1=z2=0D.-i的平方等于1【解析】選AB.對(duì)于選項(xiàng)A,因?yàn)閤,y∈R,且x+yi=1+i,依據(jù)復(fù)數(shù)相等的性質(zhì),則x=y=1,故正確;對(duì)于選項(xiàng)B,因?yàn)樘摂?shù)不能比較大小,故正確;對(duì)于選項(xiàng)C,因?yàn)槿魪?fù)數(shù)z1=i,z2=1滿意QUOTE+QUOTE=0,則z1≠z2≠0,故不正確;對(duì)于選項(xiàng)D,因?yàn)閺?fù)數(shù)QUOTE=-1,故不正確.【補(bǔ)償訓(xùn)練】(多選題)(2024·濰坊高二檢測)已知i為虛數(shù)單位,下列命題中正確的是 ()A.若x,y∈C,則x+yi=3-2i的充要條件是x=3,y=-2B.(a2+1)i(a∈R)是純虛數(shù)C.若實(shí)數(shù)a與ai對(duì)應(yīng),則實(shí)數(shù)集與純虛集一一對(duì)應(yīng)D.當(dāng)m=4時(shí),復(fù)數(shù)lg(m2-2m-7)+(m2+5m+6)i是純虛數(shù)【解析】選BD.取x=-2i,y=-3i,則x+yi=3-2i,但不滿意x=3,y=-2,故A錯(cuò)誤;?a∈R,a2+1>0恒成立,所以(a2+1)i是純虛數(shù),故B正確;若a=0,則ai不是純虛數(shù),故C錯(cuò)誤;m=4時(shí),復(fù)數(shù)lg(m2-2m-7)+(m2+5m+6)i=42i是純虛數(shù),故D正確.三、填空題(每小題5分,共10分)7.若log2(x2-3x-2)+ilog2(x2+2x+1)>1,則實(shí)數(shù)x的值是________.

【思路導(dǎo)引】復(fù)數(shù)值大于1,則復(fù)數(shù)必為實(shí)數(shù),即虛部為0,實(shí)部大于1.【解析】因?yàn)閘og2(x2-3x-2)+ilog2(x2+2x+1)>1,所以QUOTE即QUOTE解得x=-2.答案:-28.復(fù)數(shù)z=cosQUOTE+isinQUOTE,且θ∈QUOTE,若z是實(shí)數(shù),則θ的值為________;若z為純虛數(shù),則θ的值為________.

【解析】z=cosQUOTE+isinQUOTE=-sinθ+icosθ.當(dāng)z是實(shí)數(shù)時(shí),cosθ=0,因?yàn)棣取蔘UOTE,所以θ=±QUOTE;當(dāng)z為純虛數(shù)時(shí)QUOTE又θ∈QUOTE,所以θ=0.答案:±QUOTE0四、解答題(每小題10分,共20分)9.(2024·通化高二檢測)已知復(fù)數(shù)z:m(m-1)+(m2+2m-3)i,當(dāng)m取何值時(shí)復(fù)數(shù)z是:(1)實(shí)數(shù);(2)純虛數(shù);(3)z=2+5i.【解析】(1)因?yàn)閦為實(shí)數(shù),所以m2+2m-3=0,解得m=-3或m=1,所以當(dāng)m=-3或m=1時(shí),z為實(shí)數(shù);(2)由z為純虛數(shù),可得QUOTE即QUOTE解得m=0,所以當(dāng)m=0時(shí),z為純虛數(shù);(3)因?yàn)閦=2+5i,所以m(m-1)+(m2+2m-3)i=2+5i,所以QUOTE解得m=2,所以當(dāng)m=2時(shí)z=2+5i.10.已知集合M={(a+3)+(b2-1)i,8},集合N={3i,(a2-1)+(b+2)i},且M∩NM,M∩N≠?,求整數(shù)a,b的值.【解析】若M∩N={3i},則(a+3)+(b2-1)i=3i,即a+3=0且b2-1=3,即a=-3,b=±2.當(dāng)a=-3,b=-2時(shí),M={3i,8},N={3i,8},M∩N=M,不合題意,舍去;當(dāng)a=-3,b=2時(shí),M={3i,8},N={3i,8+4i},符合題意,所以a=-3,b=2.若M∩N={8},則8=(a2-1)+(b+2)i,即