新高考數(shù)學(xué)二輪復(fù)習(xí) 小題綜合練專題02 復(fù)數(shù)（解析版）

專題02復(fù)數(shù)小題綜合一、單選題1．（2023·浙江·校聯(lián)考模擬預(yù)測(cè)）若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．3 D．2【答案】A【分析】利用復(fù)數(shù)的除法運(yùn)算及求模公式計(jì)算即可.【詳解】由SKIPIF1<0，故選：A2．（2023·浙江·校聯(lián)考二模）已知復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0（SKIPIF1<0為虛數(shù)單位），則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】根據(jù)復(fù)數(shù)的乘法運(yùn)算規(guī)則計(jì)算.【詳解】SKIPIF1<0；故選：B.3．（2023·浙江·校聯(lián)考模擬預(yù)測(cè)）已知SKIPIF1<0，則在復(fù)平面內(nèi)，復(fù)數(shù)SKIPIF1<0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】C【分析】利用復(fù)數(shù)除法法則計(jì)算得到SKIPIF1<0，從而確定復(fù)數(shù)SKIPIF1<0對(duì)應(yīng)的點(diǎn)所在象限.【詳解】由SKIPIF1<0可得，SKIPIF1<0則復(fù)數(shù)SKIPIF1<0對(duì)應(yīng)的點(diǎn)為SKIPIF1<0，位于第三象限.故選：C.4．（2023·浙江·高三專題練習(xí)）設(shè)i為虛數(shù)單位，若復(fù)數(shù)z滿足SKIPIF1<0，則z的虛部為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．2【答案】D【分析】根據(jù)復(fù)數(shù)的乘、除法運(yùn)算及虛部的概念即可求解．【詳解】由SKIPIF1<0，則SKIPIF1<0，所以z的虛部為2．故選：D．5．（2023·浙江·校聯(lián)考模擬預(yù)測(cè)）已知SKIPIF1<0是虛數(shù)單位，SKIPIF1<0，SKIPIF1<0，則“SKIPIF1<0”是“SKIPIF1<0”的（

）．A．充分不必要條件 B．必要不充分條件C．充分必要條件 D．既不充分也不必要條件【答案】A【分析】根據(jù)復(fù)數(shù)的相關(guān)運(yùn)算，由充分不必要條件的概念判斷即可.【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0；反之，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0.所以SKIPIF1<0，則SKIPIF1<0，所以不一定得到SKIPIF1<0.綜上：“SKIPIF1<0”是“SKIPIF1<0”的充分不必要條件.故選：A6．（2023·浙江金華·統(tǒng)考模擬預(yù)測(cè)）若復(fù)數(shù)SKIPIF1<0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，SKIPIF1<0，由題意可得SKIPIF1<0，SKIPIF1<0，解方程即可得出答案.【詳解】設(shè)SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0.故SKIPIF1<0SKIPIF1<0.故選：C.7．（2023·浙江·校聯(lián)考模擬預(yù)測(cè)）已知復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0（

）A．5 B．SKIPIF1<0 C．13 D．SKIPIF1<0【答案】B【分析】設(shè)SKIPIF1<0，利用復(fù)數(shù)的運(yùn)算法則和復(fù)數(shù)相等，建立SKIPIF1<0的方程組，直接求出SKIPIF1<0，從而可求出結(jié)果.【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0.故選：B.8．（2023春·浙江·高三校聯(lián)考開學(xué)考試）若復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)關(guān)于虛軸對(duì)稱，且SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】由已知可推出SKIPIF1<0，然后根據(jù)復(fù)數(shù)的除法即可求出.【詳解】復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)為SKIPIF1<0，則SKIPIF1<0在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0.故選：C.9．（2023秋·浙江·高三浙江省永康市第一中學(xué)校聯(lián)考期末）若復(fù)數(shù)SKIPIF1<0滿SKIPIF1<0（SKIPIF1<0為虛數(shù)單位），則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】將已知等式SKIPIF1<0整理成SKIPIF1<0，在根據(jù)復(fù)數(shù)的除法運(yùn)算化簡(jiǎn)即可.【詳解】解：因?yàn)镾KIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0.故選：B.10．（2023秋·浙江麗水·高三浙江省麗水中學(xué)校聯(lián)考期末）設(shè)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】根據(jù)共軛復(fù)數(shù)的概念以及復(fù)數(shù)的乘法運(yùn)算，即可得答案.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，故選：B11．（2023春·浙江溫州·高三統(tǒng)考開學(xué)考試）已知SKIPIF1<0，下列選項(xiàng)中不是方程SKIPIF1<0的根的是（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】利用因式分解與復(fù)數(shù)的性質(zhì)求根即可.【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，故選項(xiàng)ACD中是方程SKIPIF1<0的根，B中不是.故選：B12．（2023秋·浙江紹興·高三統(tǒng)考期末）設(shè)復(fù)數(shù)SKIPIF1<0（SKIPIF1<0為虛數(shù)單位），則SKIPIF1<0（

）A．2 B．SKIPIF1<0 C．SKIPIF1<0 D．1【答案】D【分析】根據(jù)復(fù)數(shù)計(jì)算規(guī)則計(jì)算即可.【詳解】SKIPIF1<0，所以SKIPIF1<0；故選：D13．（2023·浙江紹興·紹興一中?？寄M預(yù)測(cè)）已知SKIPIF1<0，SKIPIF1<0是關(guān)于x的方程SKIPIF1<0的兩個(gè)根.若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．1 C．SKIPIF1<0 D．2【答案】C【分析】由SKIPIF1<0，SKIPIF1<0是關(guān)于x的方程SKIPIF1<0的兩個(gè)根，由韋達(dá)定理求出SKIPIF1<0，再由復(fù)數(shù)的模長(zhǎng)公式求解即可.【詳解】法一：由SKIPIF1<0，SKIPIF1<0是關(guān)于x的方程SKIPIF1<0的兩個(gè)根，得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.法二：由SKIPIF1<0，SKIPIF1<0是關(guān)于x的方程SKIPIF1<0的兩個(gè)根，得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：C.14．（2023·浙江·校聯(lián)考三模）已知復(fù)數(shù)SKIPIF1<0是純虛數(shù)，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．12 C．SKIPIF1<0 D．3【答案】C【分析】根據(jù)復(fù)數(shù)的除法運(yùn)算化簡(jiǎn)SKIPIF1<0，根據(jù)純虛數(shù)的概念列式計(jì)算，可得答案.【詳解】由題意SKIPIF1<0，因?yàn)閺?fù)數(shù)SKIPIF1<0是純虛數(shù)，故SKIPIF1<0，解得SKIPIF1<0，故選：C15．（2023春·浙江金華·高三浙江金華第一中學(xué)校考階段練習(xí)）已知復(fù)數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．2 B．3 C．4 D．5【答案】C【分析】對(duì)復(fù)數(shù)SKIPIF1<0去分母，將化簡(jiǎn)得到SKIPIF1<0，對(duì)應(yīng)系數(shù)相等即可得到SKIPIF1<0的值，進(jìn)而求得SKIPIF1<0的值.【詳解】SKIPIF1<0SKIPIF1<0SKIPIF1<0則SKIPIF1<0故選：C.16．（2023·浙江·統(tǒng)考二模）已知復(fù)數(shù)SKIPIF1<0（i是虛數(shù)單位），則z的虛部為（

）．A．2 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】由復(fù)數(shù)的模長(zhǎng)、乘法和除法運(yùn)算化簡(jiǎn)復(fù)數(shù)，即可得出答案.【詳解】SKIPIF1<0，故z的虛部為2.故選：A.17．（2023·浙江·高三專題練習(xí)）若復(fù)數(shù)z滿足SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】根據(jù)復(fù)數(shù)的除法運(yùn)算，化簡(jiǎn)可得SKIPIF1<0，然后根據(jù)共軛復(fù)數(shù)的概念，即可得出答案.【詳解】由已知可得，SKIPIF1<0，從而SKIPIF1<0.故選：B.18．（2023·浙江·高三專題練習(xí)）已知SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．0 C．SKIPIF1<0 D．1【答案】A【分析】利用復(fù)數(shù)的四則運(yùn)算計(jì)算求模即可.【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，解之得SKIPIF1<0，所以SKIPIF1<0.故選:A19．（2023·浙江紹興·統(tǒng)考二模）已知復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，其中SKIPIF1<0為虛數(shù)單位，則SKIPIF1<0的虛部為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】根據(jù)復(fù)數(shù)的運(yùn)算化簡(jiǎn)SKIPIF1<0，再由虛部的概念即可得答案.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0所以SKIPIF1<0的虛部為SKIPIF1<0.故選：A.20．（2023·浙江金華·統(tǒng)考模擬預(yù)測(cè)）若復(fù)數(shù)SKIPIF1<0，則復(fù)數(shù)SKIPIF1<0的模SKIPIF1<0（

）A．3 B．5 C．9 D．25【答案】B【分析】先化簡(jiǎn)求出SKIPIF1<0，再根據(jù)模長(zhǎng)公式求解即得.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：B.21．（2023·浙江紹興·統(tǒng)考模擬預(yù)測(cè)）歐拉公式SKIPIF1<0（i為虛數(shù)單位）是由瑞士著名數(shù)學(xué)家歐拉創(chuàng)立，依據(jù)歐拉公式，下列選項(xiàng)不正確的是（

）A．復(fù)數(shù)SKIPIF1<0的虛部為SKIPIF1<0 B．若SKIPIF1<0，則復(fù)數(shù)SKIPIF1<0對(duì)應(yīng)點(diǎn)位于第二象限C．復(fù)數(shù)SKIPIF1<0的模長(zhǎng)等于1 D．復(fù)數(shù)SKIPIF1<0的共軛復(fù)數(shù)為SKIPIF1<0【答案】D【分析】根據(jù)歐拉公式，即可由復(fù)數(shù)的除法運(yùn)算以及幾何意義，模長(zhǎng)公式，共軛復(fù)數(shù)的定義，結(jié)合選項(xiàng)即可求解.【詳解】SKIPIF1<0，故復(fù)數(shù)SKIPIF1<0的虛部為SKIPIF1<0，A正確，SKIPIF1<0對(duì)應(yīng)的點(diǎn)為SKIPIF1<0，由于SKIPIF1<0，所以SKIPIF1<0，故對(duì)應(yīng)的點(diǎn)為第二象限，故B正確，對(duì)于C，SKIPIF1<0，故模長(zhǎng)為SKIPIF1<0，故C正確，SKIPIF1<0，所以共軛復(fù)數(shù)為SKIPIF1<0，故D錯(cuò)誤，故選：D22．（2023·浙江嘉興·校考模擬預(yù)測(cè)）已知復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0（SKIPIF1<0是虛數(shù)單位），則SKIPIF1<0的虛部為（

）A．2 B．SKIPIF1<0 C．1 D．SKIPIF1<0【答案】B【分析】由題目條件可得SKIPIF1<0，即SKIPIF1<0，然后利用復(fù)數(shù)的運(yùn)算法則化簡(jiǎn).【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0故復(fù)數(shù)SKIPIF1<0的虛部為SKIPIF1<0.故選：B23．（2023春·浙江杭州·高三浙江省杭州第二中學(xué)?？茧A段練習(xí)）若復(fù)數(shù)z滿足SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．1【答案】C【分析】首先設(shè)復(fù)數(shù)SKIPIF1<0，(SKIPIF1<0不同時(shí)為0)，根據(jù)條件化簡(jiǎn)求得SKIPIF1<0的關(guān)系式，再根據(jù)復(fù)數(shù)模的幾何意義求最值.【詳解】設(shè)SKIPIF1<0，(SKIPIF1<0不同時(shí)為0)，SKIPIF1<0，由題意可知SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的軌跡是SKIPIF1<0軸（除原點(diǎn)外），此時(shí)SKIPIF1<0的幾何意義表示復(fù)數(shù)表示的點(diǎn)和SKIPIF1<0的距離，此時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，復(fù)數(shù)SKIPIF1<0的軌跡是以原點(diǎn)為圓心，SKIPIF1<0為半徑的圓，如圖，根據(jù)復(fù)數(shù)模的幾何意義可知，SKIPIF1<0的幾何意義是圓上的點(diǎn)到SKIPIF1<0的距離，如圖可知，SKIPIF1<0的最小值是點(diǎn)SKIPIF1<0與SK