PAGEPAGE43.1.2復(fù)數(shù)的幾何意義課時(shí)過(guò)關(guān)·實(shí)力提升基礎(chǔ)鞏固1.實(shí)部為-2,虛部為1的復(fù)數(shù)所對(duì)應(yīng)的點(diǎn)位于復(fù)平面的()A.第一象限 B.其次象限C.第三象限 D.第四象限答案:B2.復(fù)數(shù)z=3A.第一象限 B.實(shí)軸上 C.虛軸上 D.第四象限解析:因?yàn)閦=3+i所以z對(duì)應(yīng)的點(diǎn)在實(shí)軸上.故選B.答案:B3.復(fù)數(shù)z與它的模相等的充要條件是()A.z為純虛數(shù) B.z是實(shí)數(shù)C.z是正實(shí)數(shù) D.z是非負(fù)實(shí)數(shù)解析:因?yàn)閦=|z|,所以z為實(shí)數(shù),且z≥0.故選D.答案:D4.在復(fù)平面內(nèi),復(fù)數(shù)6+5i,-2+3i對(duì)應(yīng)的點(diǎn)分別為A,B.若C為線段AB的中點(diǎn),則點(diǎn)C對(duì)應(yīng)的復(fù)數(shù)是()A.4+8i B.8+2i C.2+4i D.4+i解析:由題意得點(diǎn)A(6,5),B(-2,3).由C為線段AB的中點(diǎn),得C(2,4),所以點(diǎn)C對(duì)應(yīng)的復(fù)數(shù)為2+4i.答案:C5.已知0<a<2,復(fù)數(shù)z=a+i(i是虛數(shù)單位),則|z|的取值范圍是()A.(1,3) B.(1,5) C.解析:|z|=∵0<a<2,∴0<a2<4.∴1<a2+1答案:B6.已知復(fù)數(shù)z滿意|z|2-2|z|-3=0,則復(fù)數(shù)z對(duì)應(yīng)點(diǎn)的軌跡為()A.一個(gè)圓 B.線段 C.兩點(diǎn) D.兩個(gè)圓解析:∵|z|2-2|z|-3=0,∴(|z|-3)(|z|+1)=0,∴|z|=3.故所求的軌跡為一個(gè)圓,故選A.答案:A7.復(fù)數(shù)z=cos40°-icos50°的模等于.

解析:|z|=co答案:18.若復(fù)數(shù)(m2-3m-4)+(m2-5m-6)i在復(fù)平面內(nèi)表示的點(diǎn)在虛軸上,則實(shí)數(shù)m的值等于.

解析:由題意可得m2-3m-4=0,解得m=4或m=-1.答案:4或-19.若復(fù)數(shù)z=(x-1)+(2x-1)i的模小于10,求實(shí)數(shù)x的取值范圍分析:依據(jù)復(fù)數(shù)的模的意義及題設(shè)中復(fù)數(shù)模的范圍,建立關(guān)于實(shí)數(shù)x的不等式求解即可.解:由題意,可得化簡(jiǎn)得5x2-6x-8<0,解得-故x的取值范圍是10.已知復(fù)數(shù)z=(a+3)+(2a-4)i,其中a∈R.當(dāng)復(fù)數(shù)z在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)Z滿意下列條件時(shí),求a的值(或取值范圍).(1)點(diǎn)Z在實(shí)軸上;(2)點(diǎn)Z與原點(diǎn)關(guān)于點(diǎn)(2,-1)對(duì)稱;(3)點(diǎn)Z在第四象限;(4)點(diǎn)Z在曲線x2=解:因?yàn)閦=(a+3)+(2a-4)i,所以復(fù)數(shù)z在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)Z的坐標(biāo)為(a+3,2a-4).(1)若點(diǎn)Z在實(shí)軸上,則2a-4=0,解得a=2.(2)若點(diǎn)Z與原點(diǎn)關(guān)于點(diǎn)(2,-1)對(duì)稱,則Z(4,-2),因此a+3=4,(3)若點(diǎn)Z在第四象限,則a+3>0,2(4)若點(diǎn)Z在曲線x2=12y+5上,則(a+3整理得a2+5a+6=0,解得a=-2或-3.實(shí)力提升1.若復(fù)數(shù)z1,z2在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)關(guān)于虛軸對(duì)稱,且z1=2+i,則z2=()A.2+i B.-2+iC.2-i D.-2-i答案:B2.若z=(2t2+5t-3)+(t2+2t+3)i,t∈R,則以下結(jié)論正確的是()A.z對(duì)應(yīng)的點(diǎn)在第一象限B.z肯定不是純虛數(shù)C.z肯定是純虛數(shù)D.z對(duì)應(yīng)的點(diǎn)在實(shí)軸上方解析:因?yàn)?t2+5t-3=2t2+2t+3=(t+1)2+2≥2,所以復(fù)數(shù)z對(duì)應(yīng)的點(diǎn)在實(shí)軸上方.故選D.答案:D3.使|log12x-4i|≥|3+4i|成立的A.18,8B.C.0,18∪[8,+∞) D.(0,1)∪解析:由已知得(log12x)2+(-4)2≥3∴(log12∴l(xiāng)og12x≥3或lo∴x∈0,18∪[8,+答案:C4.★已知復(fù)數(shù)3-5i,1-i和-2+ai在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)在同一條直線上,則實(shí)數(shù)a的值為()A.5 B.-2 C.-5 D.解析:設(shè)復(fù)數(shù)3-5i,1-i,-2+ai對(duì)應(yīng)的向量分別為OA,則OA=(3,-5),OB=(1,-1)∵A,B,C三點(diǎn)共線,∴OA=tOB即(3,-5)=t(1,-1)+(1-t)(-2,a).∴3=t-答案:A5.設(shè)z1=1+i,z2=-1+i,O為原點(diǎn),復(fù)數(shù)z1和z2在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)分別為A,B,則△AOB的面積為.

解析:由已知可得A(1,1),B(-1,1),則∴△AOB為等腰三角形,且AB與x軸平行,|AB|=2,∴S△AOB=12×2×答案:16.在復(fù)平面內(nèi),O是原點(diǎn),已知復(fù)數(shù)z1=-1+2i,z2=1-i,z3=3-2i,它們所對(duì)應(yīng)的點(diǎn)分別是A,B,C.若OC=xOA+yOB(x,y∈R),則解析:由已知,得OA=(-1,2),OB=(1,-1),OC=(3,-2),所以xOA+yOB=x(-1,2)答案:57.設(shè)z=log2(1+m)+ilog12(3-m)(m∈(1)若z在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)在第三象限,求m的取值范圍;(2)若z在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)在直線x-y-1=0上,求m的值.解:(1)由題意,得log2(即m的取值范圍是-1<m<0.(2)由已知得點(diǎn)(log2(1+m),log12(3-m))在直線x-y-1=0上,即log2(1+m)-log12(3-m)-1=0,所以log2[(1+m)(3-m)]=1,因此(1+m)(3-m)=2,則m2-2m-1=0,因此m=1±8.★復(fù)數(shù)z的模為1,求|z-1-i|的最大值和最小值.解:由|