人教A版高中數(shù)學(xué)必修第二冊(cè)同步講義第18講 復(fù)數(shù)的加、減運(yùn)算及其幾何意義（ 原卷版）

第18課復(fù)數(shù)的加、減運(yùn)算及其幾何意義目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航課程標(biāo)準(zhǔn)課標(biāo)解讀1.熟練掌握復(fù)數(shù)代數(shù)形式的加、減運(yùn)算法則.2.理解復(fù)數(shù)加減法的幾何意義，能夠利用“數(shù)形結(jié)合”的思想解題．1.在認(rèn)真學(xué)習(xí)復(fù)數(shù)定義的基礎(chǔ)上，熟練掌握復(fù)數(shù)代數(shù)形式的加、減運(yùn)算法則.2.進(jìn)一步加強(qiáng)理解復(fù)數(shù)加減法的幾何意義，能夠利用“數(shù)形結(jié)合”的思想解題，提升數(shù)學(xué)學(xué)科素養(yǎng).知識(shí)精講知識(shí)精講知識(shí)點(diǎn)一復(fù)數(shù)加法與減法的運(yùn)算法則1．設(shè)z1＝a＋bi，z2＝c＋di(a，b，c，d∈R)是任意兩個(gè)復(fù)數(shù)，則(1)z1＋z2＝(a＋c)＋(b＋d)i；(2)z1－z2＝.2．對(duì)任意z1，z2，z3∈C，有(1)z1＋z2＝；(2)(z1＋z2)＋z3＝．【即學(xué)即練1】設(shè)m∈R，復(fù)數(shù)z1＝eq\f(m2＋m,m＋2)＋(m－15)i，z2＝－2＋m(m－3)i，若z1＋z2是虛數(shù)，求m的取值范圍．反思感悟復(fù)數(shù)加、減運(yùn)算的解題思路兩個(gè)復(fù)數(shù)相加(減)，就是把兩個(gè)復(fù)數(shù)的實(shí)部相加(減)，虛部相加(減)．復(fù)數(shù)的減法是加法的逆運(yùn)算．當(dāng)多個(gè)復(fù)數(shù)相加(減)時(shí)，可將這些復(fù)數(shù)的所有實(shí)部相加(減)，所有虛部相加(減)．知識(shí)點(diǎn)02復(fù)數(shù)加、減法的幾何意義如圖，設(shè)復(fù)數(shù)z1，z2對(duì)應(yīng)的向量分別為eq\o(OZ1,\s\up6(→))，eq\o(OZ2,\s\up6(→))，四邊形OZ1ZZ2為平行四邊形，則向量eq\o(OZ,\s\up6(→))與復(fù)數(shù)z1＋z2對(duì)應(yīng)，向量eq\o(Z2Z1,\s\up6(→))與復(fù)數(shù)對(duì)應(yīng)．【即學(xué)即練2】若z1＝1＋2i，z2＝2＋ai，復(fù)數(shù)z2－z1所對(duì)應(yīng)的點(diǎn)在第四象限內(nèi)，則實(shí)數(shù)a的取值范圍是________．能力拓展能力拓展考法01復(fù)數(shù)代數(shù)形式的加、減運(yùn)算【典例1】復(fù)數(shù)(1＋2i)＋(3－4i)－(－5－3i)對(duì)應(yīng)的點(diǎn)在()A．第一象限 B．第二象限C．第三象限 D．第四象限考法02復(fù)數(shù)加、減法的幾何意義【典例2】如圖所示，平行四邊形OABC的頂點(diǎn)O，A，C對(duì)應(yīng)的復(fù)數(shù)分別為0,3＋2i，－2＋4i.求：(1)eq\o(AO,\s\up6(→))對(duì)應(yīng)的復(fù)數(shù)；(2)eq\o(CA,\s\up6(→))對(duì)應(yīng)的復(fù)數(shù)；(3)eq\o(OB,\s\up6(→))對(duì)應(yīng)的復(fù)數(shù)及eq\o(OB,\s\up6(→))的長(zhǎng)度．反思感悟復(fù)數(shù)與向量的對(duì)應(yīng)關(guān)系的兩個(gè)關(guān)注點(diǎn)(1)復(fù)數(shù)z＝a＋bi(a，b∈R)是與以原點(diǎn)為起點(diǎn)，Z(a，b)為終點(diǎn)的向量一一對(duì)應(yīng)的．(2)一個(gè)向量可以平移，其對(duì)應(yīng)的復(fù)數(shù)不變，但是其起點(diǎn)與終點(diǎn)所對(duì)應(yīng)的復(fù)數(shù)發(fā)生改變．【變式訓(xùn)練】已知復(fù)平面內(nèi)的向量eq\o(OA,\s\up6(→))，eq\o(AB,\s\up6(→))對(duì)應(yīng)的復(fù)數(shù)分別是－2＋i,3＋2i，則|eq\o(OB,\s\up6(→))|＝________.考法03復(fù)數(shù)模的綜合問(wèn)題【典例3】如果復(fù)數(shù)z滿足|z＋i|＋|z－i|＝2，那么|z＋i＋1|的最小值是()A．1B.eq\f(1,2)C．2D.eq\r(5)反思感悟兩個(gè)復(fù)數(shù)差的模的幾何意義(1)|z－z0|表示復(fù)數(shù)z，z0對(duì)應(yīng)的點(diǎn)之間的距離，在應(yīng)用時(shí)，要把絕對(duì)值號(hào)內(nèi)變?yōu)閮蓮?fù)數(shù)差的形式．(2)|z－z0|＝r表示以z0對(duì)應(yīng)的點(diǎn)為圓心，r為半徑的圓．(3)涉及復(fù)數(shù)模的最值問(wèn)題以及點(diǎn)的集合所表示的圖形問(wèn)題，均可從兩點(diǎn)間距離公式的復(fù)數(shù)表達(dá)形式入手進(jìn)行分析判斷，然后通過(guò)幾何方法進(jìn)行求解．【變式訓(xùn)練】△ABC的三個(gè)頂點(diǎn)所對(duì)應(yīng)的復(fù)數(shù)分別為z1，z2，z3，復(fù)數(shù)z滿足|z－z1|＝|z－z2|＝|z－z3|，則z對(duì)應(yīng)的點(diǎn)P是△ABC的()A．外心B．內(nèi)心C．重心D．垂心分層提分分層提分題組A基礎(chǔ)過(guò)關(guān)練一、單選題1．已知復(fù)數(shù)SKIPIF1<0，SKIPIF1<0和SKIPIF1<0滿足SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的最大值為（

）A．SKIPIF1<0 B．3 C．SKIPIF1<0 D．12．已知設(shè)SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．3 B．4 C．5 D．63．復(fù)數(shù)SKIPIF1<0的模為1，其中SKIPIF1<0為虛數(shù)單位，SKIPIF1<0，則這樣的SKIPIF1<0一共有（

）個(gè).A．9 B．10 C．11 D．無(wú)數(shù)4．已知SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．4 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0.故選：D6．若SKIPIF1<0，則SKIPIF1<0等于（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、填空題7．在復(fù)平面中，已知點(diǎn)SKIPIF1<0，復(fù)數(shù)SKIPIF1<0對(duì)應(yīng)的點(diǎn)分別為SKIPIF1<0，且滿足SKIPIF1<0，則SKIPIF1<0的最大值為_(kāi)__________.8．若SKIPIF1<0為虛數(shù)單位，復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最大值為_(kāi)______.9．已知SKIPIF1<0，SKIPIF1<0且z是復(fù)數(shù)，當(dāng)SKIPIF1<0的最大值為3，則SKIPIF1<0_______.10．已知復(fù)數(shù)SKIPIF1<0為虛數(shù)單位，則SKIPIF1<0_________.三、解答題11．對(duì)于一組復(fù)數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，…，SKIPIF1<0，令SKIPIF1<0，如果存在SKIPIF1<0，使得SKIPIF1<0，那么稱SKIPIF1<0是該復(fù)數(shù)組的“SKIPIF1<0復(fù)數(shù)”.（1）設(shè)SKIPIF1<0，若SKIPIF1<0是復(fù)數(shù)組SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的“SKIPIF1<0復(fù)數(shù)”，求實(shí)數(shù)SKIPIF1<0的取值范圍；（2）已知SKIPIF1<0，SKIPIF1<0，是否存在復(fù)數(shù)SKIPIF1<0使得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0均是復(fù)數(shù)組SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的“SKIPIF1<0復(fù)數(shù)”？若存在，求出所有的SKIPIF1<0，若不存在，說(shuō)明理由；（3）若SKIPIF1<0，復(fù)數(shù)組SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，…，SKIPIF1<0是否存在“SKIPIF1<0復(fù)數(shù)”？給出你的結(jié)論并說(shuō)明理由．12．已知復(fù)數(shù)SKIPIF1<0，根據(jù)以下條件分別求實(shí)數(shù)SKIPIF1<0的值或范圍.（1）SKIPIF1<0是純虛數(shù)；（2）SKIPIF1<0對(duì)應(yīng)的點(diǎn)在復(fù)平面的第二象限.題組B能力提升練一、單選題1．已知關(guān)于SKIPIF1<0的實(shí)系數(shù)一元二次方程SKIPIF1<0有兩個(gè)虛根SKIPIF1<0和SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的值為（

）A．2 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．關(guān)于復(fù)數(shù)，下列說(shuō)法正確的是（

）A．若SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0B．復(fù)數(shù)SKIPIF1<0與SKIPIF1<0分別對(duì)應(yīng)向量SKIPIF1<0與SKIPIF1<0，則向量SKIPIF1<0對(duì)應(yīng)的復(fù)數(shù)為SKIPIF1<0C．若點(diǎn)Z的坐標(biāo)為SKIPIF1<0，則Z對(duì)應(yīng)的點(diǎn)在第三象限D(zhuǎn)．若復(fù)數(shù)z滿足SKIPIF1<0，則復(fù)數(shù)z對(duì)應(yīng)的點(diǎn)所構(gòu)成的圖形面積為SKIPIF1<03．已知SKIPIF1<0，且SKIPIF1<0，其中SKIPIF1<0，SKIPIF1<0為實(shí)數(shù)，則SKIPIF1<0（

）A．1 B．3 C．SKIPIF1<0 D．54．已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．1 B．SKIPIF1<0 C．2 D．SKIPIF1<05．若|z＋3i|＝|z＋4－i|，則|z|＋|z－2|的最小值為（

）A．2 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．歐拉公式被稱為世界上最完美的公式，歐拉公式又稱為歐拉定理，是用在復(fù)分析領(lǐng)域的公式，歐拉公式將三角函數(shù)與復(fù)數(shù)指數(shù)函數(shù)相關(guān)聯(lián)，即SKIPIF1<0（SKIPIF1<0）.根據(jù)歐拉公式，下列說(shuō)法不正確的是（

）A．對(duì)任意的SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)在第二象限C．SKIPIF1<0的實(shí)部為SKIPIF1<0 D．SKIPIF1<0與SKIPIF1<0互為共軛復(fù)數(shù)二、多選題7．已知SKIPIF1<0（SKIPIF1<0，SKIPIF1<0是虛數(shù)單位），SKIPIF1<0，定義：SKIPIF1<0，則下列結(jié)論正確的是（

）A．對(duì)任意SKIPIF1<0，都有SKIPIF1<0B．若SKIPIF1<0是z的共軛復(fù)數(shù)，則SKIPIF1<0恒成立C．若SKIPIF1<0，則SKIPIF1<0D．對(duì)任意SKIPIF1<0，則SKIPIF1<0恒成立三、填空題8．若復(fù)數(shù)SKIPIF1<0和復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0_____．9．已知SKIPIF1<0，則SKIPIF1<0____．10．已知復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，則復(fù)平面內(nèi)由點(diǎn)SKIPIF1<0形成的區(qū)域的面積為_(kāi)_____．11．已知i為虛數(shù)單位，復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)關(guān)于原點(diǎn)對(duì)稱，且SKIPIF1<0，則SKIPIF1<0_______．四、解答題12．復(fù)數(shù)SKIPIF1<0，求實(shí)數(shù)m的取值范圍使得：(1)z為純虛數(shù)；(2)z在復(fù)平面上對(duì)應(yīng)的點(diǎn)在第四象限．13．已知復(fù)數(shù)SKIPIF1<0均為銳角，且SKIPIF1<0.(1)求SKIPIF1<0的值；(2)若SKIPIF1<0，求SKIPIF1<0的值.題組C培優(yōu)拔尖練一、單選題1．已知復(fù)數(shù)z滿足SKIPIF1<0，則z的虛部是（

）A．SKIPIF1<0 B．1 C．SKIPIF1<0 D．i2．設(shè)復(fù)數(shù)SKIPIF1<0在復(fù)平面上對(duì)應(yīng)向量SKIPIF1<0，將向量SKIPIF1<0繞原點(diǎn)O按順時(shí)針?lè)较蛐D(zhuǎn)SKIPIF1<0后得到向量SKIPIF1<0，SKIPIF1<0對(duì)應(yīng)復(fù)數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、多選題3．已知復(fù)數(shù)SKIPIF1<0(i為虛數(shù)單位)在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)為SKIPIF1<0