新高考數(shù)學(xué)一輪復(fù)習(xí)過關(guān)訓(xùn)練第33課 數(shù)系的擴(kuò)充與復(fù)數(shù)的引入（含解析）

試卷第=page11頁，共=sectionpages33頁試卷第=page11頁，共=sectionpages33頁第33課數(shù)系的擴(kuò)充與復(fù)數(shù)的引入學(xué)校:___________姓名：___________班級：___________考號：___________【基礎(chǔ)鞏固】1．（2022·全國·高考真題）SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】利用復(fù)數(shù)的乘法可求SKIPIF1<0.【詳解】SKIPIF1<0，故選：D.2．（2022·浙江·高考真題）已知SKIPIF1<0（SKIPIF1<0為虛數(shù)單位），則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】利用復(fù)數(shù)相等的條件可求SKIPIF1<0.【詳解】SKIPIF1<0，而SKIPIF1<0為實(shí)數(shù)，故SKIPIF1<0，故選：B.3．（2022·北京·高考真題）若復(fù)數(shù)z滿足SKIPIF1<0，則SKIPIF1<0（

）A．1 B．5 C．7 D．25【答案】B【分析】利用復(fù)數(shù)四則運(yùn)算，先求出SKIPIF1<0，再計(jì)算復(fù)數(shù)的模．【詳解】由題意有SKIPIF1<0，故SKIPIF1<0．故選：B．4．（2022·山東青島·二模）復(fù)數(shù)SKIPIF1<0（SKIPIF1<0是虛數(shù)單位）的虛部是（

）A．1 B．SKIPIF1<0 C．2 D．SKIPIF1<0【答案】A【分析】利用復(fù)數(shù)的除法法則及復(fù)數(shù)的概念即可求解.【詳解】由題意可知，SKIPIF1<0，所以復(fù)數(shù)SKIPIF1<0的虛部為SKIPIF1<0.故選：A.5．（2021·全國·高考真題）復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)對應(yīng)的點(diǎn)所在的象限為（

）A．第一象限 B．第二象限 C．第三象限 D．第四象限【答案】A【分析】利用復(fù)數(shù)的除法可化簡SKIPIF1<0，從而可求對應(yīng)的點(diǎn)的位置.【詳解】SKIPIF1<0，所以該復(fù)數(shù)對應(yīng)的點(diǎn)為SKIPIF1<0，該點(diǎn)在第一象限，故選：A.6．（2021·全國·高考真題（文））已知SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】由已知得SKIPIF1<0，根據(jù)復(fù)數(shù)除法運(yùn)算法則，即可求解.【詳解】SKIPIF1<0，SKIPIF1<0.故選：B.7．（2022·廣東茂名·二模）已知復(fù)數(shù)z在復(fù)平面內(nèi)對應(yīng)的點(diǎn)為SKIPIF1<0，SKIPIF1<0是z的共軛復(fù)數(shù)，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】求出SKIPIF1<0，再由復(fù)數(shù)的除法運(yùn)算可得答案.【詳解】∵復(fù)數(shù)z在復(fù)平面內(nèi)對應(yīng)的點(diǎn)為SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.故選：B．8．（2022·全國·高考真題（理））已知SKIPIF1<0，且SKIPIF1<0，其中a，b為實(shí)數(shù)，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】先算出SKIPIF1<0,再代入計(jì)算,實(shí)部與虛部都為零解方程組即可【詳解】SKIPIF1<0SKIPIF1<0由SKIPIF1<0,得SKIPIF1<0,即SKIPIF1<0故選:SKIPIF1<09．（2021·全國·高考真題（理））設(shè)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】設(shè)SKIPIF1<0，利用共軛復(fù)數(shù)的定義以及復(fù)數(shù)的加減法可得出關(guān)于SKIPIF1<0、SKIPIF1<0的等式，解出這兩個(gè)未知數(shù)的值，即可得出復(fù)數(shù)SKIPIF1<0.【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，所以，SKIPIF1<0，解得SKIPIF1<0，因此，SKIPIF1<0.故選：C.10．（2022·江蘇·鹽城中學(xué)模擬預(yù)測）設(shè)復(fù)數(shù)z的模長為1，在復(fù)平面對應(yīng)的點(diǎn)位于第一象限，且滿足SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】設(shè)SKIPIF1<0，且SKIPIF1<0，利用SKIPIF1<0得SKIPIF1<0，模長為1得SKIPIF1<0，求出SKIPIF1<0后可得SKIPIF1<0.【詳解】設(shè)SKIPIF1<0，因?yàn)樵趶?fù)平面對應(yīng)的點(diǎn)位于第一象限，所以SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，因?yàn)閺?fù)數(shù)z的模長為1，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0.故選：C.11．（多選）（2022·山東濰坊·二模）若復(fù)數(shù)SKIPIF1<0，SKIPIF1<0，其中SKIPIF1<0是虛數(shù)單位，則下列說法正確的是（

）A．SKIPIF1<0B．SKIPIF1<0C．若SKIPIF1<0是純虛數(shù)，那么SKIPIF1<0D．若SKIPIF1<0，SKIPIF1<0在復(fù)平面內(nèi)對應(yīng)的向量分別為SKIPIF1<0，SKIPIF1<0（O為坐標(biāo)原點(diǎn)），則SKIPIF1<0【答案】BC【分析】利用復(fù)數(shù)的運(yùn)算法則和幾何意義即可進(jìn)行判斷.【詳解】對于A，SKIPIF1<0，A錯(cuò)誤；對于B，∵SKIPIF1<0，∴SKIPIF1<0；又SKIPIF1<0，∴SKIPIF1<0，B正確；對于C，∵SKIPIF1<0為純虛數(shù)，∴SKIPIF1<0，解得：SKIPIF1<0，C正確；對于D，由題意得：SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，D錯(cuò)誤.故選：BC12．（多選）（2022·山東泰安·模擬預(yù)測）已知復(fù)數(shù)SKIPIF1<0滿足方程SKIPIF1<0，則（

）A．SKIPIF1<0可能為純虛數(shù) B．該方程共有兩個(gè)虛根C．SKIPIF1<0可能為SKIPIF1<0 D．該方程的各根之和為2【答案】ACD【分析】依題意可得SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，從而求出SKIPIF1<0，即可判斷；【詳解】解：由SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，即方程的根分別為SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，所以SKIPIF1<0故選：ACD.13．（多選）（2022·福建省福州第一中學(xué)三模）設(shè)復(fù)數(shù)SKIPIF1<0，當(dāng)a變化時(shí)，下列結(jié)論正確的是（

）A．SKIPIF1<0恒成立 B．z可能是純虛數(shù)C．SKIPIF1<0可能是實(shí)數(shù) D．SKIPIF1<0的最大值為SKIPIF1<0【答案】ABD【分析】首先根據(jù)題意得到SKIPIF1<0，再結(jié)合復(fù)數(shù)的定義和運(yùn)算性質(zhì)依次判斷選項(xiàng)即可.【詳解】SKIPIF1<0，對選項(xiàng)A，SKIPIF1<0，SKIPIF1<0，故A正確.對選項(xiàng)B，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為純虛數(shù)，故B正確.對選項(xiàng)C，SKIPIF1<0令SKIPIF1<0，即SKIPIF1<0無解，故C錯(cuò)誤.對選項(xiàng)D，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號.所以SKIPIF1<0的最大值為SKIPIF1<0，故D正確.故選：ABD14．（多選）（2022·福建·廈門一中模擬預(yù)測）已知復(fù)數(shù)SKIPIF1<0對應(yīng)的向量為SKIPIF1<0，復(fù)數(shù)SKIPIF1<0對應(yīng)的向量為SKIPIF1<0，則（

）A．若SKIPIF1<0，則SKIPIF1<0B．若SKIPIF1<0，則SKIPIF1<0C．若SKIPIF1<0與SKIPIF1<0在復(fù)平面上對應(yīng)的點(diǎn)關(guān)于實(shí)軸對稱，則SKIPIF1<0D．若SKIPIF1<0，則SKIPIF1<0【答案】ABC【分析】利用向量數(shù)量積的運(yùn)算法則及復(fù)數(shù)的幾何意義即可求解.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，故選項(xiàng)SKIPIF1<0正確；因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，故選項(xiàng)SKIPIF1<0正確；設(shè)SKIPIF1<0，因?yàn)镾KIPIF1<0與SKIPIF1<0在復(fù)平面上對應(yīng)的點(diǎn)關(guān)于實(shí)軸對稱，則SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故選項(xiàng)SKIPIF1<0正確；若SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0，而SKIPIF1<0，故選項(xiàng)SKIPIF1<0錯(cuò)誤；故選：ABC.15．（2022·天津·高考真題）已知SKIPIF1<0是虛數(shù)單位，化簡SKIPIF1<0的結(jié)果為_______．【答案】SKIPIF1<0【分析】根據(jù)復(fù)數(shù)代數(shù)形式的運(yùn)算法則即可解出．【詳解】SKIPIF1<0．故答案為：SKIPIF1<0．16．（2022·湖南岳陽·模擬預(yù)測）已知復(fù)數(shù)z滿足SKIPIF1<0，則SKIPIF1<0_____________【答案】4【分析】根據(jù)復(fù)數(shù)的運(yùn)算公式求出復(fù)數(shù)SKIPIF1<0的代數(shù)形式，再由復(fù)數(shù)模的公式求SKIPIF1<0.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故答案為：417．（2022·江蘇·華羅庚中學(xué)三模）已知復(fù)數(shù)SKIPIF1<0，則SKIPIF1<0＝________．【答案】SKIPIF1<0【分析】根據(jù)復(fù)數(shù)的乘除法與共軛復(fù)數(shù)的概念求解即可【詳解】SKIPIF1<0，故SKIPIF1<0故答案為：SKIPIF1<018．（2022·浙江省新昌中學(xué)模擬預(yù)測）若復(fù)數(shù)z滿足SKIPIF1<0，則z的模為____________，虛部為____________．【答案】

1

SKIPIF1<0【分析】根據(jù)復(fù)數(shù)的除法運(yùn)算與模的運(yùn)算，結(jié)合虛部的定義求解即可【詳解】由題，SKIPIF1<0，故SKIPIF1<0，虛部為SKIPIF1<0.故答案為：SKIPIF1<0；SKIPIF1<019．（2022·浙江·三模）中國古代數(shù)學(xué)著作《九章算術(shù)》中記載了平方差公式，平方差公式是指兩個(gè)數(shù)的和與這兩個(gè)數(shù)差的積，等于這兩個(gè)數(shù)的平方差.若復(fù)數(shù)SKIPIF1<0（i為虛數(shù)單位），則SKIPIF1<0__________．【答案】SKIPIF1<0【分析】先要平方差公式，再按照復(fù)數(shù)的四則運(yùn)算規(guī)則計(jì)算即可.【詳解】SKIPIF1<0；故答案為：SKIPIF1<0.20．（2022·全國·模擬預(yù)測）請寫出一個(gè)同時(shí)滿足①SKIPIF1<0；②SKIPIF1<0的復(fù)數(shù)z，z=______．【答案】SKIPIF1<0【分析】設(shè)SKIPIF1<0，根據(jù)模長公式得出SKIPIF1<0，進(jìn)而得出SKIPIF1<0.【詳解】設(shè)SKIPIF1<0，由條件①可以得到SKIPIF1<0，兩邊平方化簡可得SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0；故答案為：SKIPIF1<021．（2022·北京·101中學(xué)三模）設(shè)m為實(shí)數(shù)，復(fù)數(shù)SKIPIF1<0（這里i為虛數(shù)單位），若SKIPIF1<0為純虛數(shù)，則SKIPIF1<0的值為______．【答案】SKIPIF1<0【分析】先根據(jù)SKIPIF1<0為純虛數(shù)計(jì)算出SKIPIF1<0的值，再計(jì)算SKIPIF1<0，最后計(jì)算SKIPIF1<0的值【詳解】SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0為純虛數(shù)SKIPIF1<0SKIPIF1<0SKIPIF1<0故答案為：SKIPIF1<022．（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【分析】根據(jù)復(fù)數(shù)的幾何意義SKIPIF1<0表示SKIPIF1<0，SKIPIF1<0兩點(diǎn)間距離，結(jié)合圖形理解運(yùn)算．【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．【素養(yǎng)提升】1．（2022·全國·高三專題練習(xí)）已知復(fù)數(shù)SKIPIF1<0對應(yīng)的點(diǎn)在第二象限，SKIPIF1<0為SKIPIF1<0的共軛復(fù)數(shù)，有下列關(guān)于SKIPIF1<0的四個(gè)命題：甲：SKIPIF1<0；

乙：SKIPIF1<0；丙：SKIPIF1<0；

?。篠KIPIF1<0．如果只有一個(gè)假命題，則該命題是（

）A．甲 B．乙 C．丙 D．丁【答案】B【分析】設(shè)SKIPIF1<0，根據(jù)復(fù)數(shù)所在象限、復(fù)數(shù)加法、減法、乘法和除法，結(jié)合“只有一個(gè)假命題”進(jìn)行分析，由此確定正確選項(xiàng).【詳解】設(shè)SKIPIF1<0，由于SKIPIF1<0對應(yīng)點(diǎn)在第二象限，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.甲SKIPIF1<0，乙SKIPIF1<0，丙SKIPIF1<0，丁SKIPIF1<0，由于“只有一個(gè)假命題”，所以乙是假命題，SKIPIF1<0的值應(yīng)為SKIPIF1<0.故選：B2．（2022·江蘇·揚(yáng)中市第二高級中學(xué)模擬預(yù)測）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_______.【答案】SKIPIF1<0【分析】利用復(fù)數(shù)的幾何意義知復(fù)數(shù)SKIPIF1<0對應(yīng)的點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0表示復(fù)數(shù)SKIPIF1<0對應(yīng)的點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離，數(shù)形結(jié)合可求得結(jié)果.【詳解】復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，即SKIPIF1<0即復(fù)數(shù)SKIPIF1<0對應(yīng)的點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離SKIPIF1<0滿足SKIPIF1<0設(shè)SKIPIF1<0，SKIPIF1<0表示復(fù)數(shù)SKIPIF1<0對應(yīng)的點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離數(shù)形結(jié)合可知SKIPIF1<0的最大值SKIPIF1<0故答案為：SKIPIF1<03．（2022·上海市復(fù)興高級中學(xué)高三階段練習(xí)）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_______.【答案】SKIPIF1<0【分析】由SKIPIF1<0可知，SKIPIF1<0，化簡SKIPIF1<0可得其最值為SKIPIF1<0，進(jìn)而求出SKIPIF1<0的值.【詳解】設(shè)SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，解得，SKIPIF1<0，故答案為：SKIPIF1<0.4．（2022·全國·高三專題練習(xí)）若非零復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的值是___________.【答案】SKIPIF1<0【分析】由題設(shè)有SKIPIF1<0、SKIPIF1<0易得SKIPIF1<0，同理SKIPIF1<0，SKIPIF1<0，而SKIPIF1<0