高中數(shù)學(xué)必修二 第七章 7.2 7.2.1

1、7.2.1復(fù)數(shù)的加、減運(yùn)算及其幾何意義知識(shí)點(diǎn)一復(fù)數(shù)的加法與減法(1)復(fù)數(shù)的加減法運(yùn)算法則(abi)(cdi)eq o(,sup4(01)(ac)(bd)i.(2)復(fù)數(shù)加法的運(yùn)算律復(fù)數(shù)的加法滿足eq o(,sup4(02)交換律、eq o(,sup4(03)結(jié)合律，即對(duì)任何z1，z2，z3C，有z1z2eq o(,sup4(04)z2z1；(z1z2)z3eq o(,sup4(05)z1(z2z3)知識(shí)點(diǎn)二復(fù)數(shù)加、減法的幾何意義(1)復(fù)數(shù)加法的幾何意義設(shè)eq o(OZ1,sup16()，eq o(OZ2,sup16()分別與復(fù)數(shù)abi，cdi對(duì)應(yīng)，則eq o(OZ1,sup16()(a，b)，e

2、q o(OZ2,sup16()(c，d)由平面向量的坐標(biāo)運(yùn)算法則，得eq o(OZ1,sup16()eq o(OZ2,sup16()(ac，bd)這說(shuō)明兩個(gè)向量eq o(OZ1,sup16()與eq o(OZ2,sup16()的和就是與復(fù)數(shù)(ac)(bd)i對(duì)應(yīng)的向量因此復(fù)數(shù)的加法可以按照向量加法來(lái)進(jìn)行(2)復(fù)數(shù)減法的幾何意義復(fù)數(shù)z1z2是連接向量eq o(OZ1,sup16()，eq o(OZ2,sup16()的eq o(,sup4(01)終點(diǎn)，并指向被減向量的向量eq o(Z2Z1,sup16()所對(duì)應(yīng)的復(fù)數(shù)設(shè)z1x1y1i，z2x2y2i，則d|Z1Z2|eq o(Z2Z1,sup16(

3、)|z1z2|(x1y1i)(x2y2i)|(x1x2)(y1y2)i|eq r(x1x22y1y22).(3)復(fù)平面內(nèi)的兩點(diǎn)間距離公式：deq o(,sup4(02)|z1z2|.其中z1，z2是復(fù)平面內(nèi)的兩點(diǎn)Z1和Z2所對(duì)應(yīng)的復(fù)數(shù)，d為Z1和Z2間的距離如圖：設(shè)復(fù)數(shù)z1，z2對(duì)應(yīng)向量分別為eq o(OZ1,sup16()，eq o(OZ2,sup16()，四邊形OZ1ZZ2為平行四邊形，則與z1z2對(duì)應(yīng)的向量是eq o(,sup4(03)eq o(OZ,sup16()，與z1z2對(duì)應(yīng)的向量是eq o(,sup4(04)eq o(Z2Z1,sup16().復(fù)數(shù)模的兩個(gè)重要性質(zhì)(1)|z1|z

4、2|z1z2|z1|z2|；(2)|z1z2|2|z1z2|22|z1|22|z2|2.1判一判(正確的打“”，錯(cuò)誤的打“”)(1)復(fù)數(shù)與向量一一對(duì)應(yīng)()(2)復(fù)數(shù)與復(fù)數(shù)相加減后結(jié)果只能是實(shí)數(shù)()(3)因?yàn)樘摂?shù)不能比較大小，所以虛數(shù)的模也不能比較大小()(4)兩個(gè)共軛虛數(shù)的差為純虛數(shù)()答案(1)(2)(3)(4)2做一做(1)計(jì)算：(35i)(34i)_.(2)(56i)(22i)(33i)_.(3)已知向量eq o(OZ1,sup16()對(duì)應(yīng)的復(fù)數(shù)為23i，向量eq o(OZ2,sup16()對(duì)應(yīng)的復(fù)數(shù)為34i，則向量eq o(Z1Z2,sup16()對(duì)應(yīng)的復(fù)數(shù)為_答案(1)6i(2)11

5、i(3)1i題型一 復(fù)數(shù)的加、減運(yùn)算例1計(jì)算：(1)(35i)(4i)(34i)；(2)(7i5)(98i)(32i)解(1)原式(343)(514)i410i.(2)原式(593)(782)i1i.復(fù)數(shù)代數(shù)形式的加、減法運(yùn)算，其運(yùn)算法則是對(duì)它們的實(shí)部和虛部分別進(jìn)行加、減運(yùn)算在運(yùn)算過(guò)程中應(yīng)注意把握每一個(gè)復(fù)數(shù)的實(shí)部和虛部這種運(yùn)算類似于初中的合并同類項(xiàng)計(jì)算：(1)(12i)(2i)(2i)(12i)；(2)(i2i)|i|(1i)解(1)原式(13i)(2i)(12i)(32i)(12i)2.(2)原式(1i)eq r(012)(1i)1i1(1i)12i.題型二 復(fù)數(shù)加、減運(yùn)算的幾何意義例2已知

6、四邊形ABCD是復(fù)平面內(nèi)的平行四邊形，且A，B，C三點(diǎn)對(duì)應(yīng)的復(fù)數(shù)分別是13i，i,2i，求點(diǎn)D對(duì)應(yīng)的復(fù)數(shù)解解法一：設(shè)點(diǎn)D對(duì)應(yīng)的復(fù)數(shù)為xyi(x，yR)，則D(x，y)又由已知得A(1,3)，B(0，1)，C(2,1)，AC中點(diǎn)為eq blc(rc)(avs4alco1(f(3,2)，2)，BD中點(diǎn)為eq blc(rc)(avs4alco1(f(x,2)，f(y1,2).平行四邊形對(duì)角線互相平分，eq blcrc (avs4alco1(f(3,2)f(x,2)，,2f(y1,2)，)eq blcrc (avs4alco1(x3，,y5.)即點(diǎn)D對(duì)應(yīng)的復(fù)數(shù)為35i.解法二：設(shè)點(diǎn)D對(duì)應(yīng)的復(fù)數(shù)為xyi

7、(x，yR)則eq o(AD,sup16()對(duì)應(yīng)的復(fù)數(shù)為(xyi)(13i)(x1)(y3)i，又eq o(BC,sup16()對(duì)應(yīng)的復(fù)數(shù)為(2i)(i)22i.由已知得eq o(AD,sup16()eq o(BC,sup16()，(x1)(y3)i22i，eq blcrc (avs4alco1(x12，,y32，)eq blcrc (avs4alco1(x3，,y5，)即點(diǎn)D對(duì)應(yīng)的復(fù)數(shù)為35i.條件探究若一個(gè)平行四邊形的三個(gè)頂點(diǎn)對(duì)應(yīng)的復(fù)數(shù)分別為13i，i,2i，求第四個(gè)頂點(diǎn)對(duì)應(yīng)的復(fù)數(shù)解設(shè)13i，i,2i對(duì)應(yīng)A，B，C三點(diǎn)，D為第四個(gè)頂點(diǎn)，則當(dāng)四邊形ABCD是平行四邊形時(shí)，點(diǎn)D對(duì)應(yīng)的復(fù)數(shù)是35

8、i.當(dāng)四邊形ABDC是平行四邊形時(shí)，點(diǎn)D對(duì)應(yīng)的復(fù)數(shù)為13i.當(dāng)四邊形ADBC是平行四邊形時(shí)，點(diǎn)D對(duì)應(yīng)的復(fù)數(shù)為1i.(1)根據(jù)復(fù)數(shù)的兩種幾何意義可知：復(fù)數(shù)的加、減運(yùn)算可以轉(zhuǎn)化為點(diǎn)的坐標(biāo)運(yùn)算或向量運(yùn)算(2)復(fù)數(shù)的加減運(yùn)算用向量進(jìn)行時(shí)，同樣滿足平行四邊形法則和三角形法則(3)復(fù)數(shù)及其加減運(yùn)算的幾何意義為數(shù)形結(jié)合思想在復(fù)數(shù)中的應(yīng)用提供了可能已知復(fù)平面內(nèi)平行四邊形ABCD，A點(diǎn)對(duì)應(yīng)的復(fù)數(shù)為2i，向量eq o(BA,sup16()對(duì)應(yīng)的復(fù)數(shù)為12i，向量eq o(BC,sup16()對(duì)應(yīng)的復(fù)數(shù)為3i，求：(1)點(diǎn)C，D對(duì)應(yīng)的復(fù)數(shù)；(2)平行四邊形ABCD的面積解(1)因?yàn)橄蛄縠q o(BA,sup16()

9、對(duì)應(yīng)的復(fù)數(shù)為12i，向量eq o(BC,sup16()對(duì)應(yīng)的復(fù)數(shù)為3i，所以向量eq o(AC,sup16()對(duì)應(yīng)的復(fù)數(shù)為(3i)(12i)23i.又eq o(OC,sup16()eq o(OA,sup16()eq o(AC,sup16()，所以點(diǎn)C對(duì)應(yīng)的復(fù)數(shù)為(2i)(23i)42i.因?yàn)閑q o(AD,sup16()eq o(BC,sup16()，所以向量eq o(AD,sup16()對(duì)應(yīng)的復(fù)數(shù)為3i，即eq o(AD,sup16()(3，1)，設(shè)D(x，y)，則eq o(AD,sup16()(x2，y1)(3，1)，所以eq blcrc (avs4alco1(x23，,y11，)解得eq

10、 blcrc (avs4alco1(x5，,y0.)所以點(diǎn)D對(duì)應(yīng)的復(fù)數(shù)為5.(2)因?yàn)閑q o(BA,sup16()eq o(BC,sup16()|eq o(BA,sup16()|eq o(BC,sup16()|cosB，所以cosBeq f(o(BA,sup16()o(BC,sup16(),|o(BA,sup16()|o(BC,sup16()|)eq f(32,r(5)r(10)eq f(1,5r(2)eq f(r(2),10).所以sinBeq f(7,5r(2)eq f(7r(2),10)，所以S|eq o(BA,sup16()|eq o(BC,sup16()|sinBeq r(5)eq

11、 r(10)eq f(7r(2),10)7.所以平行四邊形ABCD的面積為7.題型三 復(fù)數(shù)加、減運(yùn)算的幾何意義的應(yīng)用例3已知|z1|z2|z1z2|1，求|z1z2|.解解法一：設(shè)z1abi，z2cdi(a，b，c，dR)，|z1|z2|z1z2|1，a2b2c2d21，(ac)2(bd)21.由得2ac2bd1.|z1z2|eq r(ac2bd2)eq r(a2c2b2d22ac2bd)eq r(3).解法二：設(shè)O為坐標(biāo)原點(diǎn)，z1，z2，z1z2對(duì)應(yīng)的點(diǎn)分別為A，B，C.|z1|z2|z1z2|1，OAB是邊長(zhǎng)為1的正三角形，四邊形OACB是一個(gè)內(nèi)角為60，邊長(zhǎng)為1的菱形，且|z1z2|是菱

12、形的較長(zhǎng)的對(duì)角線OC的長(zhǎng)，|z1z2|OC|eq r(|OA|2|AC|22|OA|AC|cos120)eq r(3).掌握以下常用結(jié)論：在復(fù)平面內(nèi)，z1，z2對(duì)應(yīng)的點(diǎn)為A，B，z1z2對(duì)應(yīng)的點(diǎn)為C，O為坐標(biāo)原點(diǎn)，則四邊形OACB：為平行四邊形；若|z1z2|z1z2|，則四邊形OACB為矩形；若|z1|z2|，則四邊形OACB為菱形；若|z1|z2|且|z1z2|z1z2|，則四邊形OACB為正方形若復(fù)數(shù)z滿足|zi|zi|2，求|zi1|的最小值解解法一：設(shè)復(fù)數(shù)i，i，(1i)在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)分別為Z1，Z2，Z3.如圖，因?yàn)閨zi|zi|2，|Z1Z2|2，所以復(fù)數(shù)z對(duì)應(yīng)的點(diǎn)Z的集合為

13、線段Z1Z2.問題轉(zhuǎn)化為：動(dòng)點(diǎn)Z在線段Z1Z2上移動(dòng)，求|ZZ3|的最小值，由圖可知|Z1Z3|為最小值且最小值為1.解法二：設(shè)zxyi(x，yR)因?yàn)閨zi|zi|2，所以eq r(x2y12)eq r(x2y12)2，又eq r(x2y12)2eq r(x2y12)0，所以0 eq r(x2y12)2，因?yàn)閑q r(x2y12)2eq r(x2y12)，所以兩邊平方可得1yeq r(x2y12)，即(1y)2x2(y1)2，且01y2.所以x0且1y1，則zyi(1y1)所以|zi1|1(y1)i|eq r(12y12)1，等號(hào)在y1即zi時(shí)成立所以|zi1|的最小值為1.1復(fù)數(shù)z13i，

14、z21i，則z1z2在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)位于()A第一象限 B第二象限C第三象限 D第四象限答案A解析z1z2(3i)(1i)22i，z1z2在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)位于第一象限2已知|z|3，且z3i是純虛數(shù)，則eq o(z,sup6()等于()A3i B3iC3i D4i答案A解析設(shè)zxyi(x，yR)，由z3ix(y3)i為純虛數(shù)，得x0，且y3，又|z|eq r(x2y2)|y|3，y3，z3i，eq o(z,sup6()3i.故選A.3非零復(fù)數(shù)z1，z2分別對(duì)應(yīng)復(fù)平面內(nèi)的向量Oeq o(A,sup16()，Oeq o(B,sup16()，若|z1z2|z1z2|，則()AOeq o(A,su

15、p16()Oeq o(B,sup16() B|Oeq o(A,sup16()|Oeq o(B,sup16()|COeq o(A,sup16()Oeq o(B,sup16() DOeq o(A,sup16()，Oeq o(B,sup16()共線答案C解析如圖，由向量的加法及減法法則可知，Oeq o(C,sup16()Oeq o(A,sup16()Oeq o(B,sup16()，Beq o(A,sup16()Oeq o(A,sup16()Oeq o(B,sup16().由復(fù)數(shù)加法及減法的幾何意義可知，|z1z2|對(duì)應(yīng)Oeq o(C,sup16()的模，|z1z2|對(duì)應(yīng)Beq o(A,sup16()的模又|z1z2|z1z2|，所以四邊形OACB是矩形，則Oeq o(A,sup16()Oeq o(B,sup16().4復(fù)數(shù)z滿足z(1i)2i，則z等于()A1i B1iC1i D1i答案A解析z2i(1i)1i.故選A.5如圖所示，平行四邊形OABC的頂點(diǎn)O，A，C分別對(duì)應(yīng)復(fù)數(shù)0,32i，24i.求：(1)向量eq o(AO,sup16()對(duì)應(yīng)的復(fù)數(shù)(2)向量eq o(CA,sup16()對(duì)應(yīng)的復(fù)數(shù)(3)向量eq o(OB,sup16()對(duì)應(yīng)的