2024-2025學(xué)年新教材高中數(shù)學(xué)課時(shí)素養(yǎng)檢測(cè)六復(fù)數(shù)的加法與減法含解析新人教B版必修第四冊(cè)

PAGE六復(fù)數(shù)的加法與減法(30分鐘60分)一、選擇題(每小題4分,共24分,多選題全部選對(duì)得4分,選對(duì)但不全對(duì)得2分,有選錯(cuò)的得0分)1.已知復(fù)數(shù)z1=1-i,z2=2+5i,則|z1+z2|= ()A.3 B.4 C.5 D.6【解析】選C.由z1=1-i,z2=2+5i,得z1+z2=(1-i)+(2+5i)=(1+2)+(-1+5)i=3+4i,所以|z1+z2|=5.2.已知復(fù)數(shù)z,“z+QUOTE=0”是“z為純虛數(shù)”的 ()A.充分不必要條件 B.必要不充分條件C.充要條件 D.既不充分也不必要條件【解析】選B.當(dāng)z=0時(shí),滿意z+QUOTE=0,此時(shí)為實(shí)數(shù);而當(dāng)z為純虛數(shù)時(shí),z+QUOTE=0,所以“z+QUOTE=0”是“z為純虛數(shù)”的必要不充分條件.3.如圖,在復(fù)平面內(nèi),復(fù)數(shù)z1,z2對(duì)應(yīng)的向量分別是,,則|z1+z2|等于 ()A.1 B.QUOTE C.2 D.3【解析】選B.由題干圖可知z1=-2-2i,z2=i,所以z1+z2=-2-i,所以|z1+z2|=QUOTE.4.A,B分別是復(fù)數(shù)z1,z2在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn),O是原點(diǎn),若|z1+z2|=|z1-z2|,則三角形AOB肯定是 ()A.等腰三角形 B.直角三角形C.等邊三角形 D.等腰直角三角形【解析】選B.依據(jù)復(fù)數(shù)加(減)法的幾何意義,以,為鄰邊所作的平行四邊形的對(duì)角線相等,則此平行四邊形為矩形,故三角形OAB為直角三角形.5.設(shè)復(fù)數(shù)z=x+yi,x,y是實(shí)數(shù),且x+2i=yi-1,則|z-QUOTE|= ()A.1 B.2 C.3 D.4【解析】選D.由x+2i=yi-1,得x=-1,y=2,所以z=-1+2i,QUOTE=-1-2i,|z-QUOTE|=|4i|=4.【補(bǔ)償訓(xùn)練】假如復(fù)數(shù)z=3+ai滿意條件|z-2|<2,那么實(shí)數(shù)a的取值范圍是 ()A.(-2QUOTE,2QUOTE) B.(-2,2)C.(-1,1) D.(-QUOTE,QUOTE)【解析】選D.|z-2|<2即|1+ai|<2,所以QUOTE<2,所以-QUOTE<a<QUOTE.6.(多選題)下列關(guān)于復(fù)數(shù)的敘述正確的是 ()A.兩個(gè)共軛復(fù)數(shù)的和是實(shí)數(shù)B.兩個(gè)共軛復(fù)數(shù)的差是虛數(shù)C.兩個(gè)共軛虛數(shù)的和是實(shí)數(shù)D.兩個(gè)共軛虛數(shù)的差是虛數(shù)【解析】選ACD.設(shè)復(fù)數(shù)z=a+bi,a,b∈R,則共軛復(fù)數(shù)QUOTE=a-bi,所以有z+QUOTE=2a∈R,z-QUOTE=2bi,當(dāng)b=0時(shí),z-QUOTE是實(shí)數(shù),當(dāng)b≠0時(shí),z-QUOTE是虛數(shù),所以A正確,B不正確.設(shè)虛數(shù)z=a+bi,a,b∈R,且b≠0,則共軛虛數(shù)QUOTE=a-bi,所以有z+QUOTE=2a∈R,z-QUOTE=2bi是虛數(shù),所以C正確,D正確.二、填空題(每小題4分,共8分)7.計(jì)算|(3-i)+(-1+2i)-(-1-3i)|=________.

【解析】|(3-i)+(-1+2i)-(-1-3i)|=|(2+i)-(-1-3i)|=|3+4i|=QUOTE=5.答案:58.已知|z|=3,且z+3i是純虛數(shù),則z=________.

【解析】設(shè)z=a+bi,a,b∈R,由|z|=3,得a2+b2=9,且z+3i=a+(b+3)i是純虛數(shù),得a=0,b=±3,當(dāng)z=-3i時(shí),z+3i=0,不是純虛數(shù),所以z=3i.答案:3i【補(bǔ)償訓(xùn)練】復(fù)平面內(nèi)有A,B,C三點(diǎn),點(diǎn)A對(duì)應(yīng)的復(fù)數(shù)是3+i,向量對(duì)應(yīng)的復(fù)數(shù)是-2-4i,向量對(duì)應(yīng)的復(fù)數(shù)是-4-i,則B點(diǎn)對(duì)應(yīng)的復(fù)數(shù)為________.

【解析】因?yàn)楸硎镜膹?fù)數(shù)是2+4i,表示的復(fù)數(shù)是4+i,所以=-=(4+i)-(2+4i)=2-3i,故=+=(3+i)+(2-3i)=5-2i,所以B點(diǎn)對(duì)應(yīng)的復(fù)數(shù)為zB=5-2i.答案:5-2i三、解答題(每小題14分,共28分)9.已知復(fù)數(shù)z1=1+2i,z2=-2+i,z3=-1-2i對(duì)應(yīng)的點(diǎn)是一個(gè)平行四邊形的三個(gè)頂點(diǎn),如何求這個(gè)平行四邊形的第四個(gè)頂點(diǎn)對(duì)應(yīng)的復(fù)數(shù)?請(qǐng)說明理由.【解析】設(shè)復(fù)數(shù)z1,z2,z3對(duì)應(yīng)的點(diǎn)分別為A,B,C,平行四邊形的第四個(gè)頂點(diǎn)D對(duì)應(yīng)的復(fù)數(shù)為x+yi(x,y∈R),若AC為平行四邊形的一條對(duì)角線,則=,得-=-,即(x+yi)-(1+2i)=(-1-2i)-(-2+i),(x-1)+(y-2)i=1-3i.所以QUOTE解得QUOTE故點(diǎn)D對(duì)應(yīng)的復(fù)數(shù)為2-i.若BC為平行四邊形的一條對(duì)角線,則=,同理,得點(diǎn)D對(duì)應(yīng)的復(fù)數(shù)為-4-3i.若AB為平行四邊形的一條對(duì)角線,則=,同理,得點(diǎn)D對(duì)應(yīng)的復(fù)數(shù)為5i.10.已知z1,z2∈C,|z1|=|z2|=1,|z1+z2|=QUOTE,求|z1-z2|.【解析】設(shè)復(fù)數(shù)z1,z2,z1+z2在復(fù)平面上對(duì)應(yīng)的點(diǎn)分別為Z1,Z2,Z,由|z1|=|z2|=1知,以O(shè)Z1,OZ2為鄰邊的平行四邊形是菱形,在△OZ1Z中,由余弦定理,得cos∠OZ1Z=QUOTE=-QUOTE,所以∠OZ1Z=120°,所以∠Z1OZ2=60°,因此△OZ1Z2是正三角形,所以|z1-z2|=|Z2Z1|=1.(35分鐘70分)一、選擇題(每小題4分,共16分,多選題全部選對(duì)得4分,選對(duì)但不全對(duì)得2分,有選錯(cuò)的得0分)1.已知復(fù)數(shù)z1=1+3i,z2=2-ai,(a∈R),若z1+z2的和所對(duì)應(yīng)的點(diǎn)在實(shí)軸上,則a的值為 ()A.1 B.2 C.3 D.-3【解析】選C.由復(fù)數(shù)z1=1+3i,z2=2-ai,得z1+z2=3+(3-a)i,由于對(duì)應(yīng)的點(diǎn)在實(shí)軸上,則a=3.2.在復(fù)平面上復(fù)數(shù)-3-2i,-4+5i,2+i所對(duì)應(yīng)的點(diǎn)分別是A,B,C,則平行四邊形ABCD的對(duì)角線BD所對(duì)應(yīng)的復(fù)數(shù)是 ()A.5-9i B.-5-3iC.7-11i D.-7+11i【解析】選C.在復(fù)平面上復(fù)數(shù)-3-2i,-4+5i,2+i所對(duì)應(yīng)的點(diǎn)分別是A,B,C,則=(-3,-2),=(-4,5),=(2,1),所以平行四邊形ABCD的對(duì)角線BD滿意=+=(-)+(-)=(7,-11),所對(duì)應(yīng)的復(fù)數(shù)是7-11i.【補(bǔ)償訓(xùn)練】已知復(fù)數(shù)z1=1+2i,z2=3-4i,若z+z1=z2-z,則復(fù)數(shù)z= ()A.1+3i B.1-3iC.2-6i D.3+4i【解析】選B.設(shè)z=a+bi,a,b∈R,由復(fù)數(shù)z1=1+2i,z2=3-4i,且z+z1=z2-z,得a+bi+1+2i=3-4i-(a+bi),得(a+1)+(b+2)i=(3-a)+(-4-b)i,所以a+1=3-a,b+2=-4-b,得a=1,b=-3,所以z=1-3i.【一題多解】因?yàn)閺?fù)數(shù)z1=1+2i,z2=3-4i,且z+z1=z2-z,所以2z=z2-z1=3-4i-(1+2i)=2-6i,所以z=1-3i.3.(多選題)設(shè)z1,z2∈C,則下列關(guān)系正確的是 ()A.|z1+z2|>|z1| B.|z1-z2|<|z1|C.|z1+z2|≤|z1|+|z2| D.|z1-z2|≤|z1|+|z2|【解析】選CD.若z2=0時(shí),|z1+z2|=|z1|,|z1-z2|=|z1|,故A,B不正確.設(shè)復(fù)數(shù)z1,z2對(duì)應(yīng)平面對(duì)量,,當(dāng)與不共線時(shí),|+|<||+||,當(dāng)與方向相同時(shí),|+|=||+||,故|+|≤||+||,即|z1+z2|≤|z1|+|z2|,C正確.當(dāng)與不共線時(shí),|-|<||+||,當(dāng)與方向相反時(shí),|-|=||+||,故|-|≤||+||,即|z1-z2|≤|z1|+|z2|,D正確.4.|(3+2i)-(1+i)|表示復(fù)平面上 ()A.點(diǎn)(3,2)與點(diǎn)(1,1)之間的距離B.點(diǎn)(3,2)與點(diǎn)(-1,-1)之間的距離C.點(diǎn)(3,2)到原點(diǎn)的距離D.以上都不對(duì)【解析】選A.設(shè)z1=3+2i,z2=1+i,原式=|z1-z2|,其幾何意義表示的就是這兩個(gè)復(fù)數(shù)對(duì)應(yīng)的兩個(gè)點(diǎn)之間的距離.二、填空題(每小題4分,共16分)5.復(fù)平面內(nèi)三點(diǎn)A,B,C,點(diǎn)A對(duì)應(yīng)的復(fù)數(shù)為3-4i,向量對(duì)應(yīng)的復(fù)數(shù)為1+2i,向量對(duì)應(yīng)的復(fù)數(shù)為3-2i,則點(diǎn)C對(duì)應(yīng)的復(fù)數(shù)為________.

【解析】由點(diǎn)A對(duì)應(yīng)的復(fù)數(shù)為3-4i,向量對(duì)應(yīng)的復(fù)數(shù)為1+2i,向量對(duì)應(yīng)的復(fù)數(shù)為3-2i,得=+=+-=(3-4i)+(3-2i)-(1+2i)=5-8i,所以點(diǎn)C對(duì)應(yīng)的復(fù)數(shù)為5-8i.答案:5-8i6.設(shè)復(fù)數(shù)z滿意z+|z|=2+i,則z=________.

【解析】設(shè)z=x+yi(x,y∈R),則|z|=QUOTE,所以x+yi+QUOTE=2+i,所以QUOTE解得QUOTE所以z=QUOTE+i.答案:QUOTE+i7.已知復(fù)平面上△AOB的重心G所對(duì)應(yīng)的復(fù)數(shù)為1+i,則以O(shè)A,OB為鄰邊的平行四邊形的對(duì)角線長(zhǎng)為________.

【解析】復(fù)平面上△AOB的重心G所對(duì)應(yīng)的復(fù)數(shù)為1+i,設(shè)AB的中點(diǎn)為D,則=QUOTE,由向量加法的平行四邊形法則,得+=2=3=3+3i,故以O(shè)A,OB為鄰邊的平行四邊形的對(duì)角線長(zhǎng)為3QUOTE.答案:3QUOTE8.已知|z|=QUOTE,且z-2+4i為純虛數(shù),則z=________.

【解析】設(shè)復(fù)數(shù)z=x+yi(x,y∈R),則z-2+4i=(x-2)+(y+4)i.由題意知QUOTE所以QUOTE或QUOTE所以z=2±i.答案:2±i三、解答題(共38分)9.(12分)已知復(fù)數(shù)z1=(3x+y)+(y-4x)i,z2=(4y-2x)-(5x+3y)i,(x,y∈R).設(shè)z=z1-z2=14-11i,求z1+z2.【解析】由z1=(3x+y)+(y-4x)i,z2=(4y-2x)-(5x+3y)i,(x,y∈R),z=z1-z2=14-11i得,(5x-3y)+(x+4y)i=14-11i,所以QUOTE,解得QUOTE,所以z1=(3x+y)+(y-4x)i=-7i,z2=(4y-2x)-(5x+3y)i=-14+4i,z1+z2=-14-3i.10.(12分)在平行四邊形ABCD中,已知,對(duì)應(yīng)的復(fù)數(shù)分別為z1=3+5i,z2=-1+2i.(1)求對(duì)應(yīng)的復(fù)數(shù).(2)求對(duì)應(yīng)的復(fù)數(shù).(3)求平行四邊形ABCD的面積.【解析】(1)由于=+=+,所以=-,故對(duì)應(yīng)的復(fù)數(shù)為z=z1-z2=(3+5i)-(-1+2i)=4+3i.(2)由于=-=-,所以對(duì)應(yīng)的復(fù)數(shù)為(4+3i)-(-1+2i)=5+i.(3)由(1)(2)可知在平行四邊形ABCD中,==(-1,2),==(4,3),所以cos∠DAB==QUOTE=QUOTE.因此sin∠DAB=QUO