新高考數(shù)學(xué)一輪復(fù)習(xí) 講與練第17講 復(fù)數(shù)（解析版）

第17講復(fù)數(shù)學(xué)校____________姓名____________班級____________一、知識梳理1.復(fù)數(shù)的有關(guān)概念(1)定義：一般地，當(dāng)a與b都是實數(shù)時，稱a＋bi為復(fù)數(shù).復(fù)數(shù)一般用小寫字母z表示，即z＝a＋bi(a，b∈R)，其中a稱為z的實部，b稱為z的虛部.(2)分類：滿足條件(a，b為實數(shù))復(fù)數(shù)的分類a＋bi為實數(shù)?b＝0a＋bi為虛數(shù)?b≠0a＋bi為純虛數(shù)?a＝0且b≠0(3)復(fù)數(shù)相等：a＋bi＝c＋di?a＝c且b＝d(a，b，c，d∈R).(4)共軛復(fù)數(shù)：如果兩個復(fù)數(shù)的實部相等，而虛部互為相反數(shù)，則稱這兩個復(fù)數(shù)互為共軛復(fù)數(shù)，復(fù)數(shù)z的共軛復(fù)數(shù)用z表示.(5)復(fù)數(shù)的模：向量eq\o(OZ,\s\up6(→))＝(a，b)的長度稱為復(fù)數(shù)z＝a＋bi(a，b∈R)的模(或絕對值)，復(fù)數(shù)z的模用|z|表示，因此|z|＝eq\r(a2＋b2).當(dāng)b＝0時，|z|＝eq\r(a2)＝|a|.2.復(fù)數(shù)的幾何意義復(fù)數(shù)z＝a＋bi與復(fù)平面內(nèi)的點(diǎn)Z(a，b)及平面向量eq\o(OZ,\s\up6(→))＝(a，b)(a，b∈R)是一一對應(yīng)關(guān)系.3.復(fù)數(shù)的運(yùn)算(1)運(yùn)算法則：設(shè)z1＝a＋bi，z2＝c＋di，a，b，c，d∈R.(2)幾何意義：復(fù)數(shù)加、減法可按向量的平行四邊形或三角形法則進(jìn)行.如圖給出的平行四邊形OZ1ZZ2可以直觀地反映出復(fù)數(shù)加、減法的幾何意義，即eq\o(OZ,\s\up6(→))＝eq\o(OZ1,\s\up6(→))＋eq\o(OZ2,\s\up6(→))，eq\o(Z1Z2,\s\up6(→))＝eq\o(OZ2,\s\up6(→))－eq\o(OZ1,\s\up6(→)).(3)由復(fù)數(shù)加、減法的幾何意義可得||z1|－|z2||≤|z1±z2|≤|z1|＋|z2|.考點(diǎn)和典型例題1、復(fù)數(shù)的概念及幾何意義【典例1-1】（2022·江西萍鄉(xiāng)·三模（理））在復(fù)平面內(nèi)，復(fù)數(shù)SKIPIF1<0所對應(yīng)的點(diǎn)關(guān)于虛軸對稱，若SKIPIF1<0，則復(fù)數(shù)SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】因為SKIPIF1<0對應(yīng)的點(diǎn)為SKIPIF1<0，SKIPIF1<0所對應(yīng)的點(diǎn)關(guān)于虛軸對稱，所以SKIPIF1<0對應(yīng)的點(diǎn)為SKIPIF1<0，所以SKIPIF1<0.故選：B.【典例1-2】（2022·江西師大附中三模（理））對任意復(fù)數(shù)SKIPIF1<0，SKIPIF1<0為虛數(shù)單位，SKIPIF1<0是z的共軛復(fù)數(shù)，則下列結(jié)論中不正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】對于A，由SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，故A正確；對于B，因為SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故B錯誤；對于C，由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，故C正確；對于D，因為SKIPIF1<0，故D正確.故選：B.【典例1-3】（2022·浙江·效實中學(xué)模擬預(yù)測）設(shè)SKIPIF1<0是虛數(shù)單位，復(fù)數(shù)SKIPIF1<0為實數(shù)，則實數(shù)SKIPIF1<0的值為（

）A．2 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】依題意，復(fù)數(shù)SKIPIF1<0為實數(shù)，所以SKIPIF1<0.故選：C【典例1-4】（2022·廣東廣州·三模）若復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，則在復(fù)平面內(nèi)SKIPIF1<0的共軛復(fù)數(shù)對應(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【詳解】由SKIPIF1<0得SKIPIF1<0，則SKIPIF1<0，則復(fù)平面內(nèi)SKIPIF1<0的共軛復(fù)數(shù)對應(yīng)的點(diǎn)位于第一象限.故選：A.【典例1-5】（2022·江蘇·南京師大附中模擬預(yù)測）設(shè)i是虛數(shù)單位，復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，則復(fù)數(shù)SKIPIF1<0的共軛復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)對應(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【答案】D【詳解】由題意得SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0，其對應(yīng)的點(diǎn)SKIPIF1<0在第四象限，故選：D【典例1-6】（2022·內(nèi)蒙古·滿洲里市教研培訓(xùn)中心三模（文））復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的虛部為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因為SKIPIF1<0，所以SKIPIF1<0，所以復(fù)數(shù)SKIPIF1<0的虛部為SKIPIF1<0，故選：A.【典例1-7】（2022·浙江省杭州學(xué)軍中學(xué)模擬預(yù)測）在復(fù)平面內(nèi)，復(fù)數(shù)SKIPIF1<0對應(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】C【詳解】由題得SKIPIF1<0，即為復(fù)平面的點(diǎn)SKIPIF1<0，故在第三象限.故選：C．【典例1-8】（2022·天津·二模）如果復(fù)數(shù)z滿足SKIPIF1<0，那么SKIPIF1<0的最大值是______．【答案】2SKIPIF1<0＃＃SKIPIF1<0＋2【詳解】設(shè)復(fù)數(shù)z在復(fù)平面中對應(yīng)的點(diǎn)為SKIPIF1<0∵SKIPIF1<0，則點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離為2，即點(diǎn)SKIPIF1<0的軌跡為以SKIPIF1<0為圓心，半徑為2的圓SKIPIF1<0表示點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離，結(jié)合圖形可得SKIPIF1<0故答案為：SKIPIF1<0．【典例1-9】（2021·上海市七寶中學(xué)模擬預(yù)測）若純虛數(shù)SKIPIF1<0滿足SKIPIF1<0，則實數(shù)SKIPIF1<0等于_________.【答案】1【詳解】解：因為SKIPIF1<0，所以SKIPIF1<0，因為SKIPIF1<0為純虛數(shù)，所以SKIPIF1<0，解得SKIPIF1<0；故答案為：SKIPIF1<0【典例1-10】（2022·天津和平·二模）復(fù)數(shù)：滿足SKIPIF1<0（SKIPIF1<0是虛數(shù)單位），則復(fù)數(shù)z在復(fù)平面內(nèi)所表示的點(diǎn)的坐標(biāo)為___________.【答案】SKIPIF1<0【詳解】由題意得：SKIPIF1<0，SKIPIF1<0SKIPIF1<0對應(yīng)的點(diǎn)的坐標(biāo)為SKIPIF1<0.故答案為：SKIPIF1<02、復(fù)數(shù)的運(yùn)算【典例2-1】（2022·江西·臨川一中模擬預(yù)測（理））已知SKIPIF1<0，SKIPIF1<0是z的共軛復(fù)數(shù)，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】由已知可得SKIPIF1<0，因此，SKIPIF1<0.故選：B.【典例2-2】（2022·江西師大附中三模（文））已知SKIPIF1<0是虛數(shù)單位，則SKIPIF1<0的虛部是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．SKIPIF1<0【答案】D【詳解】SKIPIF1<0，故其虛部為SKIPIF1<0，故選：D.【典例2-3】（2022·四川·成都市錦江區(qū)嘉祥外國語高級中學(xué)模擬預(yù)測（理））已知i為虛數(shù)單位，則復(fù)數(shù)SKIPIF1<0的虛部為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0,所以SKIPIF1<0的虛部為SKIPIF1<0.故選：A.【典例2-4】（2022·全國·模擬預(yù)測）已知復(fù)數(shù)SKIPIF1<0，i為虛數(shù)單位，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：A．【典例2-5】（2022·湖南·長沙一中模擬預(yù)測）已知復(fù)數(shù)SKIPIF1<0，SKIPIF1<0是SKIPIF1<0的共軛復(fù)數(shù)，則SKIPIF1<0（

）A．0 B．SKIPIF1<0 C．1 D．2【答案】B【詳解】∵SKIPIF1<0，所以SKIPIF1<0.故選：B．【典例2-6】（2022·山東·煙臺二中模擬預(yù)測）已知復(fù)數(shù)z滿足SKIPIF1<0（i為虛數(shù)單位），則z的共軛復(fù)數(shù)SKIPIF1<0=（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因為SKIPIF1<0所以SKIPIF1<0.故選：A【典例2-7】（2022·陜西·西北工業(yè)大學(xué)附屬中學(xué)模擬預(yù)測（理））復(fù)數(shù)SKIPIF1<0的共軛復(fù)數(shù)為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0,所以復(fù)數(shù)SKIPIF1<0的共軛復(fù)數(shù)為SKIPIF1<0.故選：A.【典例2-8】（2022·吉林長春·模擬預(yù)測（理））若SKIPIF1<0，則SKIPIF1<0（

）A．1 B．SKIPIF1<0 C．2 D．4【答案】A【詳解】SKIPIF1<0,SKIPIF1<0,SKIPIF1<0，所以SKIPIF1<0.故選：A【典例2-9】（2022·上?！つM預(yù)測）若SKIPIF1<0（i是虛數(shù)單位）是關(guān)于