新高考數(shù)學(xué)一輪復(fù)習(xí)講義 第26講 復(fù)數(shù)（含解析）

第26講復(fù)數(shù)（精講）題型目錄一覽①?gòu)?fù)數(shù)的有關(guān)概念②復(fù)數(shù)的四則運(yùn)算③復(fù)數(shù)的模長(zhǎng)④復(fù)數(shù)相等和共軛復(fù)數(shù)⑤復(fù)數(shù)的幾何意義⑥復(fù)數(shù)的三角形式一、知識(shí)點(diǎn)梳理一、知識(shí)點(diǎn)梳理一、復(fù)數(shù)的概念=1\*GB3①?gòu)?fù)數(shù)的概念：形如a＋bi(a，b∈R)的數(shù)叫做復(fù)數(shù)，a，b分別是它的實(shí)部和虛部，SKIPIF1<0叫虛數(shù)單位，滿足SKIPIF1<0（1）當(dāng)且僅當(dāng)b＝0時(shí)，a＋bi為實(shí)數(shù)；（2）當(dāng)b≠0時(shí)，a＋bi為虛數(shù)；（3）當(dāng)a＝0且b≠0時(shí)，a＋bi為純虛數(shù)．其中，兩個(gè)實(shí)部相等，虛部互為相反數(shù)的復(fù)數(shù)互為共軛復(fù)數(shù)．=2\*GB3②兩個(gè)復(fù)數(shù)SKIPIF1<0相等SKIPIF1<0（兩復(fù)數(shù)對(duì)應(yīng)同一點(diǎn)）=3\*GB3③復(fù)數(shù)的模：復(fù)數(shù)SKIPIF1<0的模，其計(jì)算公式SKIPIF1<0二、復(fù)數(shù)的加、減、乘、除的運(yùn)算法則1、復(fù)數(shù)運(yùn)算（1）SKIPIF1<0（2）SKIPIF1<0SKIPIF1<0其中SKIPIF1<0，叫z的模；SKIPIF1<0是SKIPIF1<0的共軛復(fù)數(shù)SKIPIF1<0．（3）SKIPIF1<0．實(shí)數(shù)的全部運(yùn)算律（加法和乘法的交換律、結(jié)合律、分配律及整數(shù)指數(shù)冪運(yùn)算法則）都適用于復(fù)數(shù)．2、復(fù)數(shù)的幾何意義（1）復(fù)數(shù)SKIPIF1<0對(duì)應(yīng)平面內(nèi)的點(diǎn)SKIPIF1<0；（2）復(fù)數(shù)SKIPIF1<0對(duì)應(yīng)平面向量SKIPIF1<0；（3）復(fù)平面內(nèi)實(shí)軸上的點(diǎn)表示實(shí)數(shù)，除原點(diǎn)外虛軸上的點(diǎn)表示虛數(shù)，各象限內(nèi)的點(diǎn)都表示復(fù)數(shù)．（4）復(fù)數(shù)SKIPIF1<0的模SKIPIF1<0表示復(fù)平面內(nèi)的點(diǎn)SKIPIF1<0到原點(diǎn)的距離．三、復(fù)數(shù)的三角形式（1）復(fù)數(shù)的三角表示式一般地，任何一個(gè)復(fù)數(shù)SKIPIF1<0都可以表示成SKIPIF1<0形式，其中SKIPIF1<0是復(fù)數(shù)SKIPIF1<0的模；SKIPIF1<0是以SKIPIF1<0軸的非負(fù)半軸為始邊，向量SKIPIF1<0所在射線（射線SKIPIF1<0）為終邊的角，叫做復(fù)數(shù)SKIPIF1<0的輻角．SKIPIF1<0叫做復(fù)數(shù)SKIPIF1<0的三角表示式，簡(jiǎn)稱三角形式．（2）輻角的主值任何一個(gè)不為零的復(fù)數(shù)的輻角有無(wú)限多個(gè)值，且這些值相差SKIPIF1<0的整數(shù)倍．規(guī)定在SKIPIF1<0范圍內(nèi)的輻角SKIPIF1<0的值為輻角的主值．通常記作SKIPIF1<0，即SKIPIF1<0．復(fù)數(shù)的代數(shù)形式可以轉(zhuǎn)化為三角形式，三角形式也可以轉(zhuǎn)化為代數(shù)形式．（3）三角形式下的兩個(gè)復(fù)數(shù)相等兩個(gè)非零復(fù)數(shù)相等當(dāng)且僅當(dāng)它們的模與輻角的主值分別相等．（4）復(fù)數(shù)三角形式的乘法運(yùn)算①兩個(gè)復(fù)數(shù)相乘，積的模等于各復(fù)數(shù)的模的積，積的輻角等于各復(fù)數(shù)的輻角的和，即SKIPIF1<0．（5）復(fù)數(shù)三角形式的除法運(yùn)算兩個(gè)復(fù)數(shù)相除，商的模等于被除數(shù)的模除以除數(shù)的模所得的商，商的輻角等于被除數(shù)的輻角減去除數(shù)的輻角所得的差，即SKIPIF1<0．【常用結(jié)論】①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．②SKIPIF1<0．二、題型分類精講二、題型分類精講題型一復(fù)數(shù)的有關(guān)概念策略方法解決復(fù)數(shù)概念問(wèn)題的方法及注意事項(xiàng)(1)求一個(gè)復(fù)數(shù)的實(shí)部與虛部，只需將已知的復(fù)數(shù)化為代數(shù)形式z＝a＋bi(a，b∈R)，則該復(fù)數(shù)的實(shí)部為a，虛部為b．(2)復(fù)數(shù)是實(shí)數(shù)的條件：①z＝a＋bi∈R?b＝0(a，b∈R)；②z∈R?z＝eq\x\to(z)；③z∈R?z2≥0．(3)復(fù)數(shù)是純虛數(shù)的條件：①z＝a＋bi是純虛數(shù)?a＝0且b≠0(a，b∈R)；②z是純虛數(shù)?z＋eq\x\to(z)＝0(z≠0)；③z是純虛數(shù)?z2＜0．【典例1】（單選題）已知i為虛數(shù)單位，若復(fù)數(shù)SKIPIF1<0是純虛數(shù)，則實(shí)數(shù)a等于（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．2【答案】D【分析】根據(jù)復(fù)數(shù)的乘法運(yùn)算求得復(fù)數(shù)z，根據(jù)純虛數(shù)的概念列式計(jì)算，即得答案.【詳解】由題意得SKIPIF1<0，因?yàn)樗鼮榧兲摂?shù)，所以SKIPIF1<0，解得SKIPIF1<0，故選：D.【題型訓(xùn)練】一、單選題1．（2023春·貴州黔東南·高三校考階段練習(xí)）復(fù)數(shù)SKIPIF1<0的虛部為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．2 D．16【答案】C【分析】利用虛數(shù)單位的性質(zhì)可求SKIPIF1<0，故可求其虛部.【詳解】因?yàn)镾KIPIF1<0，故SKIPIF1<0，故SKIPIF1<0的虛部為2，故選：C.2．（2023秋·廣東惠州·高三統(tǒng)考階段練習(xí)）已知復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的虛部是（

）A．2 B．2i C．1 D．i【答案】C【分析】根據(jù)復(fù)數(shù)的運(yùn)算化簡(jiǎn)SKIPIF1<0，再根據(jù)虛部的定義求解.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0,所以SKIPIF1<0的虛部是1.故選:C.3．（2023·湖南·校聯(lián)考模擬預(yù)測(cè)）復(fù)數(shù)z滿足SKIPIF1<0，則z的實(shí)部是（

）A．－1 B．1 C．－3 D．3【答案】C【分析】利用復(fù)數(shù)的四則運(yùn)算可得SKIPIF1<0，即可知z的實(shí)部是SKIPIF1<0.【詳解】由SKIPIF1<0可得SKIPIF1<0，所以z的實(shí)部是SKIPIF1<0．故選：C4．（2023·遼寧遼陽(yáng)·統(tǒng)考二模）復(fù)數(shù)SKIPIF1<0，則復(fù)數(shù)SKIPIF1<0的實(shí)部和虛部分別是（

）A．3，2 B．3，2i C．1，2 D．1，2i【答案】C【分析】應(yīng)用復(fù)數(shù)乘法運(yùn)算化簡(jiǎn)復(fù)數(shù)，即可確定實(shí)部、虛部.【詳解】由題意SKIPIF1<0，則復(fù)數(shù)SKIPIF1<0的實(shí)部和虛部分別是1和2.故選：C5．（2023·陜西安康·陜西省安康中學(xué)?？寄M預(yù)測(cè)）設(shè)復(fù)數(shù)SKIPIF1<0的實(shí)部與虛部互為相反數(shù)，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．2 D．3【答案】D【分析】根據(jù)復(fù)數(shù)的乘法運(yùn)算化簡(jiǎn)復(fù)數(shù)z，根據(jù)實(shí)部與虛部互為相反數(shù)列式計(jì)算，即得答案.【詳解】SKIPIF1<0，由已知得SKIPIF1<0，解得SKIPIF1<0，故選：D6．（2023·江蘇無(wú)錫·輔仁高中?？寄M預(yù)測(cè)）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：C7．（2023·江蘇鹽城·鹽城中學(xué)?？寄M預(yù)測(cè)）若復(fù)數(shù)SKIPIF1<0是純虛數(shù)，則SKIPIF1<0（

）A．-2 B．2 C．-1 D．1【答案】D【分析】根據(jù)復(fù)數(shù)的特征，設(shè)SKIPIF1<0（SKIPIF1<0），再根據(jù)復(fù)數(shù)的運(yùn)算，利用復(fù)數(shù)相等，列式求解.【詳解】由題意設(shè)SKIPIF1<0（SKIPIF1<0），SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，解得：SKIPIF1<0.故選：D題型二復(fù)數(shù)的四則運(yùn)算策略方法復(fù)數(shù)代數(shù)形式運(yùn)算問(wèn)題的解題策略(1)復(fù)數(shù)的加、減、乘法：復(fù)數(shù)的加、減、乘法類似于多項(xiàng)式的運(yùn)算，可將含有虛數(shù)單位i的看作一類同類項(xiàng)，不含i的看作另一類同類項(xiàng)，分別合并即可．(2)復(fù)數(shù)的除法：除法的關(guān)鍵是分子分母同乘以分母的共軛復(fù)數(shù)，使分母實(shí)數(shù)化．解題中要注意把i的冪寫成最簡(jiǎn)形式．【典例1】（單選題）若復(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】A【分析】對(duì)已知等式化簡(jiǎn)直接求解復(fù)數(shù)SKIPIF1<0【詳解】由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，故選：A【題型訓(xùn)練】一、單選題1．（2023春·湖南邵陽(yáng)·高三統(tǒng)考學(xué)業(yè)考試）若復(fù)數(shù)SKIPIF1<0（SKIPIF1<0是虛數(shù)單位），則z=（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】根據(jù)復(fù)數(shù)乘法法則計(jì)算出結(jié)果.【詳解】SKIPIF1<0.故選：B2．（2023·陜西商洛·鎮(zhèn)安中學(xué)校考模擬預(yù)測(cè)）已知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】D【分析】根據(jù)復(fù)數(shù)除法法則直接計(jì)算.【詳解】由題意得，SKIPIF1<0.故選：D3．（2023·全國(guó)·高三專題練習(xí)）SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】由復(fù)數(shù)的除法運(yùn)算即可得出答案.【詳解】SKIPIF1<0．故選:D.4．（2023春·重慶萬(wàn)州·高三重慶市萬(wàn)州第二高級(jí)中學(xué)校考階段練習(xí)）若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】根據(jù)復(fù)數(shù)的四則運(yùn)算求解即可.【詳解】由SKIPIF1<0得，SKIPIF1<0，所以SKIPIF1<0.故選：A.5．（2023·內(nèi)蒙古赤峰·赤峰二中校聯(lián)考模擬預(yù)測(cè)）若復(fù)數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)復(fù)數(shù)的運(yùn)算即可求解.【詳解】因?yàn)镾KIPIF1<0，則SKIPIF1<0，故選：C.6．（2023·全國(guó)·高三專題練習(xí)）若復(fù)數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】由復(fù)數(shù)的除法運(yùn)算即可得出答案.【詳解】SKIPIF1<0．故選:C.7．（2023·新疆喀什·?？寄M預(yù)測(cè)）已知SKIPIF1<0，則SKIPIF1<0（

）.A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．0【答案】B【分析】根據(jù)SKIPIF1<0即可得到SKIPIF1<0的值，進(jìn)而可以用復(fù)數(shù)的四則運(yùn)算法則進(jìn)行計(jì)算.【詳解】SKIPIF1<0，所以SKIPIF1<0，故選：B8．（2023·福建泉州·校聯(lián)考模擬預(yù)測(cè)）若復(fù)數(shù)SKIPIF1<0所對(duì)應(yīng)的點(diǎn)在第四象限，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】根據(jù)題意求出SKIPIF1<0，再根據(jù)復(fù)數(shù)SKIPIF1<0所對(duì)應(yīng)的點(diǎn)所在象限，即可求解.【詳解】因?yàn)閺?fù)數(shù)SKIPIF1<0滿足：SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0或SKIPIF1<0，因?yàn)閺?fù)數(shù)SKIPIF1<0所對(duì)應(yīng)的點(diǎn)在第四象限，故復(fù)數(shù)SKIPIF1<0，所以SKIPIF1<0.故選:C.題型三復(fù)數(shù)的模長(zhǎng)策略方法SKIPIF1<0【典例1】（單選題）已知復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．5【答案】B【分析】先由SKIPIF1<0化簡(jiǎn)計(jì)算求出復(fù)數(shù)SKIPIF1<0，從而可求出其模.【詳解】由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，故選：B【題型訓(xùn)練】一、單選題1．（2023春·河北石家莊·高三石家莊二中?？茧A段練習(xí)）已知復(fù)數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．2 C．SKIPIF1<0 D．10【答案】C【分析】由復(fù)數(shù)的乘法公式和模的計(jì)算公式即可求解.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0.故選：C.2．（2023秋·山西大同·高三統(tǒng)考階段練習(xí)）已知復(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】D【分析】利用復(fù)數(shù)的除法運(yùn)算求出SKIPIF1<0，再求出模作答.【詳解】依題意，SKIPIF1<0，所以SKIPIF1<0.故選：D3．（2023·重慶沙坪壩·重慶八中校考二模）若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．2 D．10【答案】A【分析】根據(jù)復(fù)數(shù)的乘法運(yùn)算及求模公式得解.【詳解】SKIPIF1<0，所以SKIPIF1<0，故選：A．4．（2023·湖南長(zhǎng)沙·周南中學(xué)?？级＃┤魪?fù)數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．4 D．5【答案】D【分析】先化簡(jiǎn)SKIPIF1<0，再由復(fù)數(shù)的加法運(yùn)算求出SKIPIF1<0，由復(fù)數(shù)的模長(zhǎng)公式求解即可.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0所以SKIPIF1<0，所以SKIPIF1<0.故選：D.5．（2023·山東菏澤·山東省鄄城縣第一中學(xué)校考三模）已知SKIPIF1<0為虛數(shù)單位，且復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0（

）A．1 B．2 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)復(fù)數(shù)的除法、乘方運(yùn)算求出SKIPIF1<0，再根據(jù)共軛復(fù)數(shù)的概念和模長(zhǎng)公式可求出結(jié)果.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0.故選：D6．（2023·四川·校聯(lián)考模擬預(yù)測(cè)）SKIPIF1<0（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．2【答案】B【分析】先根據(jù)復(fù)數(shù)得除法運(yùn)算求出復(fù)數(shù)，再根據(jù)復(fù)數(shù)的模的計(jì)算公式即可得解.【詳解】由SKIPIF1<0，得SKIPIF1<0.故選：B.7．（2023·黑龍江·黑龍江實(shí)驗(yàn)中學(xué)?？既＃┮阎獜?fù)數(shù)SKIPIF1<0滿足SKIPIF1<0（其中SKIPIF1<0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】C【分析】根據(jù)復(fù)數(shù)代數(shù)形式的除法運(yùn)算法則化簡(jiǎn)，再根據(jù)復(fù)數(shù)的定義判斷即可.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以復(fù)數(shù)SKIPIF1<0的虛部為SKIPIF1<0.故選：C8．（2023·廣東東莞·統(tǒng)考模擬預(yù)測(cè)）復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】根據(jù)復(fù)數(shù)模的公式及復(fù)數(shù)的運(yùn)算法則求得SKIPIF1<0，利用共軛復(fù)數(shù)的概念得出答案.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0．故選：A．9．（2023·全國(guó)·高三專題練習(xí)）已知復(fù)數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．10 C．SKIPIF1<0 D．2【答案】A【分析】化簡(jiǎn)復(fù)數(shù)，再由復(fù)數(shù)模長(zhǎng)公式即可得出答案.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：A.10．（2023·湖北武漢·統(tǒng)考三模）設(shè)復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0為純虛數(shù)，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】設(shè)復(fù)數(shù)SKIPIF1<0的代數(shù)形式，根據(jù)復(fù)數(shù)的除法運(yùn)算化簡(jiǎn)復(fù)數(shù)SKIPIF1<0，根據(jù)純虛數(shù)的概念以及復(fù)數(shù)的模長(zhǎng)公式可求出結(jié)果.【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，依題意得SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0.故選：A11．（2023·江蘇鹽城·統(tǒng)考三模）已知SKIPIF1<0，SKIPIF1<0，虛數(shù)SKIPIF1<0是方程SKIPIF1<0的根，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．2 D．SKIPIF1<0【答案】B【分析】將虛數(shù)z代入方程，利用復(fù)數(shù)相等解方程組即可得出答案.【詳解】因?yàn)樘摂?shù)SKIPIF1<0（SKIPIF1<0）是方程SKIPIF1<0的根，則SKIPIF1<0，即SKIPIF1<0，由復(fù)數(shù)相等得出SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，因?yàn)樘摂?shù)SKIPIF1<0中SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：B12．（2023·福建漳州·統(tǒng)考模擬預(yù)測(cè)）復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．2【答案】B【分析】根據(jù)復(fù)數(shù)的模長(zhǎng)公式即可化簡(jiǎn)求解.【詳解】設(shè)SKIPIF1<0,由SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，故選：B題型四復(fù)數(shù)相等和共軛復(fù)數(shù)策略方法解決與集合的新定義有關(guān)問(wèn)題的一般思路(1)在只含有z的方程中，z類似于代數(shù)方程中的x，可直接求解；(2)在z，eq\x\to(z)，|z|中至少含有兩個(gè)的復(fù)數(shù)方程中，可設(shè)z＝a＋bi，a，b∈R，變換方程，利用兩復(fù)數(shù)相等的充要條件得出關(guān)于a，b的方程組，求出a，b，從而得出復(fù)數(shù)z．(3)求一個(gè)復(fù)數(shù)的共軛復(fù)數(shù)，只需將此復(fù)數(shù)整理成標(biāo)準(zhǔn)的代數(shù)形式，實(shí)部不變，虛部變?yōu)橄喾磾?shù)，即得原復(fù)數(shù)的共軛復(fù)數(shù)．復(fù)數(shù)z1＝a＋bi與z2＝c＋di共軛?a＝c，b＝－d(a，b，c，d∈R)．【典例1】（單選題）已知SKIPIF1<0為虛數(shù)單位，復(fù)數(shù)SKIPIF1<0，其中a，SKIPIF1<0，則（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0 C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0【答案】B【分析】根據(jù)復(fù)數(shù)表達(dá)的唯一性求解.【詳解】SKIPIF1<0，SKIPIF1<0，故選：B.【典例2】（單選題）若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】由復(fù)數(shù)的除法和共軛復(fù)數(shù)的定義求解.【詳解】若SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0.故選：D【題型訓(xùn)練】一、單選題1．（2023·陜西咸陽(yáng)·武功縣普集高級(jí)中學(xué)?？寄M預(yù)測(cè)）已知復(fù)數(shù)SKIPIF1<0，若SKIPIF1<0的共軛復(fù)數(shù)為SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．5 C．SKIPIF1<0 D．10【答案】B【分析】利用復(fù)數(shù)運(yùn)算法則和模長(zhǎng)的性質(zhì)計(jì)算即可.【詳解】SKIPIF1<0.故選：B2．（2023·陜西西安·校考模擬預(yù)測(cè)）已知SKIPIF1<0，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．0 C．1 D．2【答案】C【分析】由復(fù)數(shù)相等的充要條件可得SKIPIF1<0的值.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，由復(fù)數(shù)相等的充要條件得SKIPIF1<0，所以SKIPIF1<0.故選：C.3．（2023·四川成都·四川省成都列五中學(xué)?？既＃┮阎獜?fù)數(shù)z滿足SKIPIF1<0，則SKIPIF1<0（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．2【答案】B【分析】運(yùn)用復(fù)數(shù)乘法運(yùn)算及復(fù)數(shù)相等可求得a、b的值，再運(yùn)用共軛復(fù)數(shù)及復(fù)數(shù)的模的運(yùn)算公式即可求得結(jié)果.【詳解】設(shè)SKIPIF1<0（a，SKIPIF1<0），則SKIPIF1<0，根據(jù)復(fù)數(shù)相等的定義，得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0．故選：B.4．（2023春·云南昆明·高三昆明一中校考階段練習(xí)）已知復(fù)數(shù)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．2【答案】B【分析】先根據(jù)復(fù)數(shù)除法法則化簡(jiǎn)復(fù)數(shù)SKIPIF1<0，代入計(jì)算即可求解.【詳解】因?yàn)镾KIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：B.5．（2023·山西大同·統(tǒng)考模擬預(yù)測(cè)）復(fù)數(shù)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)的四則運(yùn)算，求出SKIPIF1<0，再根據(jù)共軛復(fù)數(shù)的定義，即可得出SKIPIF1<0．【詳解】SKIPIF1<0．則SKIPIF1<0，故選：B．6．（2023春·安徽阜陽(yáng)·高三安徽省臨泉第一中學(xué)?？紝ｎ}練習(xí)）已知復(fù)數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】根據(jù)題意得到SKIPIF1<0，結(jié)合復(fù)數(shù)的運(yùn)算法則，即可求解.【詳解】因?yàn)镾KIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0.故選：A.7．（2023·全國(guó)·高三專題練習(xí)）已知i是虛數(shù)單位，設(shè)復(fù)數(shù)z的共軛復(fù)數(shù)為SKIPIF1<0，則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)運(yùn)算法則求SKIPIF1<0，在求其共軛復(fù)數(shù)即可.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，故選：D.8．（2023·甘肅金昌·永昌縣第一高級(jí)中學(xué)統(tǒng)考模擬預(yù)測(cè)）若復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，其中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】C【分析】設(shè)復(fù)數(shù)SKIPIF1<0，則SKIPIF1<0，根據(jù)復(fù)數(shù)的加減法與復(fù)數(shù)相等求得結(jié)果.【詳解】設(shè)復(fù)數(shù)SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0．故選：C.9．（2023·江西鷹潭·貴溪市實(shí)驗(yàn)中學(xué)校考模擬預(yù)測(cè)）已知復(fù)數(shù)SKIPIF1<0，則SKIPIF1<0的虛部為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)復(fù)數(shù)的除法運(yùn)算化簡(jiǎn)復(fù)數(shù)SKIPIF1<0，進(jìn)而求其共軛復(fù)數(shù)，即可求解.【詳解】SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0的虛部為SKIPIF1<0，故選：D.10．（2023·江西·統(tǒng)考模擬預(yù)測(cè)）已知i為虛數(shù)單位，若復(fù)數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】由復(fù)數(shù)的運(yùn)算化簡(jiǎn)復(fù)數(shù)SKIPIF1<0，再求共軛復(fù)數(shù)即可.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0.故選：B.11．（2023·全國(guó)·高三專題練習(xí)）已知SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．0 D．1【答案】A【分析】根據(jù)復(fù)數(shù)的除法運(yùn)算求出SKIPIF1<0，再由共軛復(fù)數(shù)的概念得到SKIPIF1<0，從而解出．【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0．故選：A．12．（2023·陜西咸陽(yáng)·統(tǒng)考模擬預(yù)測(cè)）若SKIPIF1<0，其中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)的相等求得SKIPIF1<0的值，再根據(jù)復(fù)數(shù)的模的計(jì)算求得答案.【詳解】由SKIPIF1<0可得SKIPIF1<0，故SKIPIF1<0，故選：B13．（2023·全國(guó)·高三專題練習(xí)）已知復(fù)數(shù)SKIPIF1<0是復(fù)數(shù)SKIPIF1<0的共軛復(fù)數(shù)，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．4 D．2【答案】C【分析】化簡(jiǎn)結(jié)合已知可得SKIPIF1<0，即可得出SKIPIF1<0的值，進(jìn)而得出答案.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：C.14．（2023·廣東廣州·廣州市從化區(qū)從化中學(xué)校考模擬預(yù)測(cè)）已知復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的共軛復(fù)數(shù)的虛部為（

）A．2 B．SKIPIF1<0 C．4 D．SKIPIF1<0【答案】B【分析】設(shè)SKIPIF1<0，根據(jù)復(fù)數(shù)的模的公式及相等復(fù)數(shù)的定義求出參數(shù)，再根據(jù)共軛復(fù)數(shù)的定義及虛部的定義即可得解.【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0的共軛復(fù)數(shù)的虛部為SKIPIF1<0.故選：B.15．（2023·江西·江西師大附中校考三模）已知復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0（SKIPIF1<0為虛數(shù)單位），則復(fù)數(shù)SKIPIF1<0的虛部為（

）A．3 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】設(shè)SKIPIF1<0，則SKIPIF1<0，代入SKIPIF1<0，利用復(fù)數(shù)相等求解.【詳解】解：設(shè)SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，所以復(fù)數(shù)SKIPIF1<0的虛部為3，故選：A16．（2023·山東煙臺(tái)·統(tǒng)考三模）已知復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．2【答案】D【分析】設(shè)SKIPIF1<0，代入SKIPIF1<0，利用復(fù)數(shù)相等求解.【詳解】解：設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0，故選：D.17．（2023·河南鄭州·統(tǒng)考模擬預(yù)測(cè)）已知SKIPIF1<0（a，SKIPIF1<0，i為虛數(shù)單位），則復(fù)數(shù)SKIPIF1<0（

）A．2 B．SKIPIF1<0 C．SKIPIF1<0 D．6【答案】B【分析】由復(fù)數(shù)的乘法運(yùn)算結(jié)合復(fù)數(shù)相等的定義求出SKIPIF1<0，SKIPIF1<0，再由模長(zhǎng)公式得出SKIPIF1<0.【詳解】∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0．故選:B．18．（2023春·湖南·高三校聯(lián)考階段練習(xí)）復(fù)數(shù)SKIPIF1<0，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】根據(jù)共軛復(fù)數(shù)、復(fù)數(shù)的乘方及復(fù)數(shù)的模一一計(jì)算可得.【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，故A錯(cuò)誤；SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故B正確；SKIPIF1<0，故C錯(cuò)誤；又SKIPIF1<0，SKIPIF1<0所以SKIPIF1<0，故D錯(cuò)誤.故選：B.題型五復(fù)數(shù)的幾何意義策略方法與復(fù)數(shù)幾何意義相關(guān)的問(wèn)題的一般解法【典例1】在復(fù)平面中，復(fù)數(shù)SKIPIF1<0（SKIPIF1<0為虛數(shù)單位）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】D【分析】利用復(fù)數(shù)的除法化簡(jiǎn)所求復(fù)數(shù)，利用復(fù)數(shù)的幾何意義可得出結(jié)論.【詳解】因?yàn)镾KIPIF1<0，該復(fù)數(shù)在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)位于第四象限.故選：D.【題型訓(xùn)練】一、單選題1．（2023·黑龍江齊齊哈爾·齊齊哈爾市實(shí)驗(yàn)中學(xué)?？既＃┮阎獜?fù)數(shù)SKIPIF1<0，其中SKIPIF1<0為虛數(shù)單位，則復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)所對(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】D【分析】先根據(jù)復(fù)數(shù)的乘法運(yùn)算求出SKIPIF1<0，再根據(jù)復(fù)數(shù)的幾何意義即可得解.【詳解】由SKIPIF1<0，可得復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)所對(duì)應(yīng)的點(diǎn)所在的象限為第四象限．故選：D．2．（2023秋·四川內(nèi)江·高三期末）復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)對(duì)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)的運(yùn)算可化簡(jiǎn)SKIPIF1<0，從而可求對(duì)應(yīng)的點(diǎn)的位置.【詳解】SKIPIF1<0，所以該復(fù)數(shù)對(duì)應(yīng)的點(diǎn)為SKIPIF1<0，該點(diǎn)在第一象限，故選：A.3．（2023·江蘇·金陵中學(xué)校聯(lián)考三模）已知復(fù)數(shù)z滿足SKIPIF1<0，則復(fù)數(shù)z在復(fù)平面內(nèi)所對(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】D【分析】利用復(fù)數(shù)除法求出z，即可判斷.【詳解】因?yàn)镾KIPIF1<0，所以點(diǎn)SKIPIF1<0位于第四象限.故選：D.4．（2023·廣東汕頭·統(tǒng)考三模）已知復(fù)數(shù)z的共軛復(fù)數(shù)SKIPIF1<0，則復(fù)數(shù)z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】D【分析】由復(fù)數(shù)的運(yùn)算求出復(fù)數(shù)z，再由復(fù)數(shù)幾何意義即可解答.【詳解】由題意SKIPIF1<0，所以SKIPIF1<0，則復(fù)數(shù)z在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)SKIPIF1<0，為第四象限內(nèi)的點(diǎn).故選：D5．（2023春·廣東茂名·高三統(tǒng)考階段練習(xí)）已知SKIPIF1<0，則復(fù)數(shù)z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【分析】根據(jù)復(fù)數(shù)四則運(yùn)算化簡(jiǎn)復(fù)數(shù)z，然后由復(fù)數(shù)的幾何意義可得.【詳解】因?yàn)镾KIPIF1<0，所以復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)為SKIPIF1<0，位于第一象限．故選：A6．（2023·河南開封·?？寄M預(yù)測(cè)）已知復(fù)數(shù)SKIPIF1<0（SKIPIF1<0是虛數(shù)單位），則SKIPIF1<0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】C【分析】先根據(jù)復(fù)數(shù)的除法運(yùn)算得到SKIPIF1<0，從而得到SKIPIF1<0，再由復(fù)數(shù)的幾何意義即可求解.【詳解】由題意得：SKIPIF1<0，所以SKIPIF1<0，由復(fù)數(shù)的幾何意義得：SKIPIF1<0在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)的坐標(biāo)為SKIPIF1<0，位于第三象限，故選：C.7．（2023·陜西西安·陜西師大附中?？寄M預(yù)測(cè)）已知復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0（其中SKIPIF1<0為虛數(shù)單位），則復(fù)數(shù)SKIPIF1<0在復(fù)平面上對(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】B【分析】根據(jù)題意化簡(jiǎn)得到SKIPIF1<0，結(jié)合復(fù)數(shù)的幾何意義，即可求解.【詳解】由復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，可得SKIPIF1<0，所以復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)為SKIPIF1<0位于第二象限.故選：B.8．（2023·吉林通化·梅河口市第五中學(xué)校考模擬預(yù)測(cè)）若復(fù)數(shù)SKIPIF1<0，則復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)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【答案】D【分析】化簡(jiǎn)z，后由復(fù)數(shù)的坐標(biāo)表示可得答案.【詳解】SKIPIF1<0，則SKIPIF1<0，則z的坐標(biāo)表示為SKIPIF1<0，則復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)在第四象限.故選：D9．（2023·河南·襄城高中校聯(lián)考三模）若復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)位于第二象限，則實(shí)數(shù)m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】化簡(jiǎn)復(fù)數(shù)為SKIPIF1<0，結(jié)合題意，列出不等式組，即可求解.【詳解】由題得SKIPIF1<0，因?yàn)閦對(duì)應(yīng)的點(diǎn)位于第二象限，所以SKIPIF1<0，解得SKIPIF1<0．故選：A.10．（2023·湖南常德·常德市一中?？寄M預(yù)測(cè)）已知復(fù)數(shù)SKIPIF1<0與SKIPIF1<0在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)關(guān)于實(shí)軸對(duì)稱，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)復(fù)數(shù)對(duì)應(yīng)點(diǎn)的對(duì)稱關(guān)系得SKIPIF1<0，應(yīng)用復(fù)數(shù)除法化簡(jiǎn)目標(biāo)式即得結(jié)果.【詳解】由SKIPIF1<0對(duì)應(yīng)點(diǎn)為SKIPIF1<0，則SKIPIF1<0對(duì)應(yīng)點(diǎn)為SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0.故選：D11．（2023·重慶萬(wàn)州·統(tǒng)考模擬預(yù)測(cè)）已知SKIPIF1<0，則復(fù)數(shù)z在復(fù)平面上對(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【答案】A【分析】設(shè)SKIPIF1<0，根據(jù)復(fù)數(shù)相等得到方程，解出SKIPIF1<0，再根據(jù)復(fù)數(shù)的幾何意義即可得到答案.【詳解】設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0,故復(fù)數(shù)SKIPIF1<0在復(fù)平面上對(duì)應(yīng)點(diǎn)SKIPIF1<0在第一象限，故選：A.12．（2023·湖北武漢·華中師大一附中?？寄M預(yù)測(cè)）復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)為SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】利用復(fù)數(shù)的幾何意義計(jì)算即可.【詳解】由SKIPIF1<0在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)為SKIPIF1<0可得SKIPIF1<0，又SKIPIF1<0SKIPIF1<0，即SKIPIF1<0.故選：D．13．（2023·河南洛陽(yáng)·洛寧縣第一高級(jí)中學(xué)?？寄M預(yù)測(cè)）在復(fù)平面內(nèi)，復(fù)數(shù)SKIPIF1<0對(duì)應(yīng)的點(diǎn)為SKIPIF1<0，則SKIPIF1<0（

）A．2 B．1 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】利用復(fù)數(shù)的幾何意義及復(fù)數(shù)的除法法則，結(jié)合復(fù)數(shù)的模公式即可求解.【詳解】因?yàn)閺?fù)數(shù)z在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)為SKIPIF1<0，所以SKIPIF1<0.所以SKIPIF1<0，所以SKIPIF1<0.故選：B.14．（2023·河北唐山·唐山市第十中學(xué)?？寄M預(yù)測(cè)）已知SKIPIF1<0，其中a，b為實(shí)數(shù)，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】D【分析】先求得在復(fù)平面內(nèi)復(fù)數(shù)SKIPIF1<0對(duì)應(yīng)的點(diǎn)的坐標(biāo)，進(jìn)而求得其所在象限.【詳解】由SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0，解之得SKIPIF1<0，則SKIPIF1<0，其在復(fù)平面內(nèi)對(duì)應(yīng)點(diǎn)的坐標(biāo)為SKIPIF1<0，該點(diǎn)位于第四象限.故選：D題型六復(fù)數(shù)的三角形式策略方法一般地，任何一個(gè)復(fù)數(shù)SKIPIF1<0都可以表示成SKIPIF1<0形式，其中SKIPIF1<0是復(fù)數(shù)SKIPIF1<0的模；SKIPIF1<0是以SKIPIF1<0軸的非負(fù)半軸為始邊，向量SKIPIF1<0所在射線（射線SKIPIF1<0）為終邊的角，叫做復(fù)數(shù)SKIPIF1<0的輻角．SKIPIF1<0叫做復(fù)數(shù)SKIPIF1<0的三角表示式，簡(jiǎn)稱三角形式．【典例1】（單選題）把復(fù)數(shù)SKIPIF1<0化三角形式為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)復(fù)數(shù)的三角形公式SKIPIF1<0求解求解即可.【詳解】設(shè)復(fù)數(shù)的三角形式為SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,可取SKIPIF1<0,從而復(fù)數(shù)SKIPIF1<0的三角形式為SKIPIF1<0.故選:C.【題型訓(xùn)練】一、單選題1．（2023·全國(guó)·高三專題練習(xí)）歐拉公式SKIPIF1<0（e為自然對(duì)數(shù)的底數(shù)，SKIPIF1<0為虛數(shù)單位）由瑞士數(shù)學(xué)家Euler（歐拉）首先發(fā)現(xiàn).它將指數(shù)函數(shù)的定義域擴(kuò)大到復(fù)數(shù)，建立了三角函數(shù)和指數(shù)函數(shù)的關(guān)系，被稱為“數(shù)學(xué)中的天橋”，則SKIPIF1<0（

）A．-1 B．1 C．-SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】根據(jù)題已知中歐拉公式SKIPIF1<0，直接計(jì)算可得答案.【詳解】由題意得：SKIPIF1<0，故選：A2．（2023·全國(guó)·高三專題練習(xí)）復(fù)數(shù)SKIPIF1<0的輻角主值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】設(shè)出輻角為SKIPIF1<0，利用公式計(jì)算出SKIPIF1<0，SKIPIF1<0，結(jié)合輻角主值的取值范圍求出答案.【詳解】設(shè)復(fù)數(shù)SKIPIF1<0的輻角為SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，滿足要求，SKIPIF1<0所以輻角主值為SKIPIF1<0.故選：A3．（2023春·廣東揭陽(yáng)·高三校考階段練習(xí)）歐拉是SKIPIF1<0世紀(jì)數(shù)學(xué)界最杰出的人物之一，他不但為數(shù)學(xué)界作出貢獻(xiàn)，更把數(shù)學(xué)推至幾乎整個(gè)物理領(lǐng)域，其中歐拉公式的諸多公式中，SKIPIF1<0(SKIPIF1<0為自然對(duì)數(shù)的底數(shù)，SKIPIF1<0為虛數(shù)單位)被稱為“數(shù)學(xué)中的天橋”，將復(fù)數(shù)?指數(shù)函數(shù)?三角函數(shù)聯(lián)系起來(lái)了.當(dāng)SKIPIF1<0時(shí)，可得恒等式（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】直接把SKIPIF1<0代入即可得．【詳解】把SKIPIF1<0代入可得SKIPIF1<