新高考數(shù)學(xué)一輪復(fù)習(xí)講與練第10講 復(fù)數(shù)（練）（解析版）

第02講復(fù)數(shù)1．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】根據(jù)復(fù)數(shù)的四則運(yùn)算，先求出復(fù)數(shù)z,再求SKIPIF1<0即可.【詳解】解：由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0.故選：C.2．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】C【分析】由復(fù)數(shù)的除法法則求解．【詳解】由SKIPIF1<0，得SKIPIF1<0.故選：C.3．若復(fù)數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0B．復(fù)數(shù)SKIPIF1<0在復(fù)平面上對應(yīng)的點在第二象限C．復(fù)數(shù)SKIPIF1<0的實部與虛部之積為SKIPIF1<0D．SKIPIF1<0【答案】A【分析】根據(jù)復(fù)數(shù)的運(yùn)算法則，化簡得到SKIPIF1<0，結(jié)合復(fù)數(shù)的基本概念，共軛復(fù)數(shù)的概念，以及復(fù)數(shù)的模的計算公式，逐項判定，即可求解.【詳解】由題意，復(fù)數(shù)SKIPIF1<0，可得SKIPIF1<0，所以A正確；復(fù)數(shù)SKIPIF1<0在復(fù)平面對應(yīng)的點SKIPIF1<0位于第三象限，所以B錯誤；復(fù)數(shù)SKIPIF1<0的實部為SKIPIF1<0，虛部為SKIPIF1<0，可得實部與虛部之積為SKIPIF1<0，所以C錯誤；由復(fù)數(shù)SKIPIF1<0的共軛復(fù)數(shù)為SKIPIF1<0，所以D錯誤.故選：A.4．若復(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)共軛復(fù)數(shù)的定義，求出SKIPIF1<0，再將SKIPIF1<0轉(zhuǎn)化為復(fù)數(shù)的標(biāo)準(zhǔn)形式即可.【詳解】由題意，SKIPIF1<0，SKIPIF1<0，∴其虛部為SKIPIF1<0；故選：D.5．已知復(fù)數(shù)SKIPIF1<0（SKIPIF1<0為虛數(shù)單位）在復(fù)平面內(nèi)對應(yīng)的點在第三象限，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】利用復(fù)數(shù)代數(shù)形式的乘除運(yùn)算化簡，再由實部與虛部均小于0列不等式組求解.【詳解】因為SKIPIF1<0，在復(fù)平面內(nèi)對應(yīng)的點在第三象限，SKIPIF1<0，解得SKIPIF1<0.故選：A.6．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】B【分析】根據(jù)復(fù)數(shù)的運(yùn)算法則，求得SKIPIF1<0，結(jié)合模的計算公式，即可求解.【詳解】由題意，復(fù)數(shù)SKIPIF1<0，所以SKIPIF1<0.故選：B.二、填空題7．已知復(fù)數(shù)SKIPIF1<0SKIPIF1<0為純虛數(shù)，則SKIPIF1<0______.【答案】4【分析】由復(fù)數(shù)為純虛數(shù)求得SKIPIF1<0的值，然后代入模的計算公式得答案．【詳解】因為復(fù)數(shù)z為純虛數(shù)，則SKIPIF1<0，解得SKIPIF1<0.所以SKIPIF1<0，所以SKIPIF1<0.故答案為：4.8．規(guī)定運(yùn)算SKIPIF1<0，若SKIPIF1<0，設(shè)SKIPIF1<0為虛數(shù)單位，則復(fù)數(shù)SKIPIF1<0__________.【答案】SKIPIF1<0【分析】根據(jù)新定義運(yùn)算直接列方程求解.【詳解】因為規(guī)定運(yùn)算SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，故答案為：SKIPIF1<09．若復(fù)數(shù)SKIPIF1<0為實數(shù)，則實數(shù)SKIPIF1<0________．【答案】SKIPIF1<0【分析】根據(jù)復(fù)數(shù)代數(shù)形式的乘方及加法運(yùn)算化簡，再根據(jù)復(fù)數(shù)的類型求出參數(shù)的值.【詳解】解：因為SKIPIF1<0為實數(shù)，所以SKIPIF1<0，即SKIPIF1<0．故答案為：SKIPIF1<010．若實數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0_____________．【答案】SKIPIF1<0#SKIPIF1<0【分析】根據(jù)復(fù)數(shù)相等充要條件，列出方程組，求得SKIPIF1<0的值，即可求解.【詳解】因為SKIPIF1<0，可得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<011．已知SKIPIF1<0，SKIPIF1<0，則“SKIPIF1<0”是“SKIPIF1<0”的________條件．【答案】充分不必要【分析】根據(jù)充分條件，必要條件的定義即得.【詳解】當(dāng)SKIPIF1<0時，必有SKIPIF1<0且SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，顯然“SKIPIF1<0”是“SKIPIF1<0”的充分不必要條件．故答案為：充分不必要.一、單選題1．已知復(fù)數(shù)z滿足SKIPIF1<0，則實數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】設(shè)SKIPIF1<0，由復(fù)數(shù)相等，得出SKIPIF1<0的關(guān)系式，消去SKIPIF1<0得到關(guān)于SKIPIF1<0的一元二次方程有實數(shù)解，利用SKIPIF1<0，求解即可得出答案.【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，整理得：SKIPIF1<0,所以SKIPIF1<0，消去SKIPIF1<0得SKIPIF1<0,因為方程有解，所以SKIPIF1<0，解得：SKIPIF1<0.故選：D.2．設(shè)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)的除法及加法法則，結(jié)合復(fù)數(shù)的摸公式即可求解.【詳解】SKIPIF1<0，所以SKIPIF1<0.故選：A.3．已知虛數(shù)z是關(guān)于x的方程SKIPIF1<0的一個根，且SKIPIF1<0，則SKIPIF1<0（

）A．1 B．2 C．4 D．5【答案】D【分析】設(shè)SKIPIF1<0，代入原方程，根據(jù)復(fù)數(shù)相等和SKIPIF1<0可得答案.【詳解】設(shè)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0），代入原方程可得SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0.故選：D.4．已知復(fù)數(shù)z的實部為1，且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,由SKIPIF1<0列方程，即可求出b，進(jìn)而得到復(fù)數(shù)z.【詳解】由題意可設(shè)：SKIPIF1<0,則SKIPIF1<0.因為SKIPIF1<0，所以SKIPIF1<0，解得：SKIPIF1<0.所以SKIPIF1<0SKIPIF1<0.故選：C5．若復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，則下列說法正確的是（

）A．SKIPIF1<0的虛部為SKIPIF1<0 B．SKIPIF1<0的共軛復(fù)數(shù)為SKIPIF1<0C．SKIPIF1<0對應(yīng)的點在第二象限 D．SKIPIF1<0【答案】C【分析】根據(jù)已知條件及復(fù)數(shù)的除法法則，再利用復(fù)數(shù)的概念及共軛復(fù)數(shù)，結(jié)合復(fù)數(shù)的幾何意義及復(fù)數(shù)的摸公式即可求解.【詳解】由SKIPIF1<0，得SKIPIF1<0，對于A，復(fù)數(shù)SKIPIF1<0的虛部為SKIPIF1<0，故A不正確；對于B，復(fù)數(shù)SKIPIF1<0的共軛復(fù)數(shù)為SKIPIF1<0，故B不正確；對于C，復(fù)數(shù)SKIPIF1<0對應(yīng)的點為SKIPIF1<0，所以復(fù)數(shù)SKIPIF1<0對應(yīng)的點在第二象限，故C正確；對于D，SKIPIF1<0，故D不正確.故選：C.6．已知命題SKIPIF1<0：SKIPIF1<0的虛部為SKIPIF1<0；命題SKIPIF1<0：在復(fù)平面內(nèi)，復(fù)數(shù)SKIPIF1<0對應(yīng)的點位于第二象限.則下列命題為真命題的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】由復(fù)數(shù)的除法和乘法運(yùn)算化簡復(fù)數(shù)，再由復(fù)數(shù)的概念和幾何意義可判斷命題SKIPIF1<0SKIPIF1<0的真假，再對各個選項進(jìn)行判斷，即可得出答案.【詳解】SKIPIF1<0，其虛部為SKIPIF1<0，命題SKIPIF1<0正確.SKIPIF1<0，在復(fù)平面內(nèi)對應(yīng)的點的坐標(biāo)為SKIPIF1<0，位于第三象限，命題SKIPIF1<0錯誤.故命題SKIPIF1<0為真命題.故選:C.7．已知復(fù)數(shù)SKIPIF1<0，若SKIPIF1<0，則當(dāng)SKIPIF1<0時，實數(shù)m的取值范圍是______________．【答案】SKIPIF1<0【分析】先對已知式子化簡計算出復(fù)數(shù)SKIPIF1<0，從而可得SKIPIF1<0，復(fù)數(shù)SKIPIF1<0，代入SKIPIF1<0中化簡可得SKIPIF1<0，從而可求出實數(shù)m的取值范圍.【詳解】SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0．由SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0．故答案為：SKIPIF1<08．已知SKIPIF1<0，則SKIPIF1<0____________．【答案】SKIPIF1<0【分析】利用復(fù)數(shù)四則運(yùn)算法則，計算SKIPIF1<0，然后利用復(fù)數(shù)相等，得SKIPIF1<0，得答案.【詳解】SKIPIF1<0，所以SKIPIF1<0，從而SKIPIF1<0．故答案為：SKIPIF1<0.9．設(shè)SKIPIF1<0（x，SKIPIF1<0），若SKIPIF1<0，則SKIPIF1<0的取值范圍是________．【答案】SKIPIF1<0【分析】根據(jù)復(fù)數(shù)的幾何意義可得復(fù)數(shù)SKIPIF1<0對應(yīng)的點的軌跡方程為圓，再轉(zhuǎn)化為圓上的點到定點的距離的最值問題即可得解.【詳解】解：由SKIPIF1<0，可得SKIPIF1<0，表示SKIPIF1<0在以SKIPIF1<0為圓心，2為半徑的圓上，SKIPIF1<0，SKIPIF1<0的幾何意義表示復(fù)平面內(nèi)點SKIPIF1<0與點SKIPIF1<0的距離，即圓SKIPIF1<0圓上的點與點SKIPIF1<0的距離，圓心SKIPIF1<0到點SKIPIF1<0的距離為SKIPIF1<0，由圓的幾何意義得到范圍是SKIPIF1<0．故答案為：SKIPIF1<0.10．復(fù)數(shù)SKIPIF1<0為純虛數(shù)，則SKIPIF1<0___________.【答案】SKIPIF1<0【分析】解不等式組SKIPIF1<0即得解.【詳解】解：∵SKIPIF1<0為純虛數(shù)，∴SKIPIF1<0解得SKIPIF1<0∴SKIPIF1<0.故答案為：SKIPIF1<011．已知復(fù)數(shù)z滿足SKIPIF1<0，則SKIPIF1<0的取值范圍是______.【答案】SKIPIF1<0【分析】根據(jù)復(fù)數(shù)模的幾何意義判斷復(fù)數(shù)z對應(yīng)點的軌跡，數(shù)形結(jié)合法判斷SKIPIF1<0的范圍.【詳解】由SKIPIF1<0，則z在復(fù)平面內(nèi)對應(yīng)的點Z是以SKIPIF1<0為圓心，2為半徑的圓上及圓內(nèi)，所以SKIPIF1<0表示Z到SKIPIF1<0的距離，故其范圍為SKIPIF1<0.故答案為：SKIPIF1<0.1．（2022·全國（理））已知SKIPIF1<0，且SKIPIF1<0，其中a，b為實數(shù)，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】先算出SKIPIF1<0,再代入計算,實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<02．（2022·全國（文））設(shè)SKIPIF1<0，其中SKIPIF1<0為實數(shù)，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】根據(jù)復(fù)數(shù)代數(shù)形式的運(yùn)算法則以及復(fù)數(shù)相等的概念即可解出．【詳解】因為SKIPIF1<0R，SKIPIF1<0，所以SKIPIF1<0，解得：SKIPIF1<0．故選：A.3．（2022·全國（文））若SKIPIF1<0．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)代數(shù)形式的運(yùn)算法則，共軛復(fù)數(shù)的概念以及復(fù)數(shù)模的計算公式即可求出．【詳解】因為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0．故選：D.4．（2022·全國（理））若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】由共軛復(fù)數(shù)的概念及復(fù)數(shù)的運(yùn)算即可得解.【詳解】SKIPIF1<0SKIPIF1<0故選：C5．（2022·全國）若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．2【答案】D【分析】利用復(fù)數(shù)的除法可求SKIPIF1<0，從而可求SKIPIF1<0.【詳解】由題設(shè)有SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0，故選：D6．（2022·全國）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.7．（2022·北京）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，再計算復(fù)數(shù)的模．【詳解】由題意有SKIPIF1<0，故SKIPIF1<0．故選：B．8．（2022·浙江）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為實數(shù)，故SKIPIF1<0，故選：B.9．（2021·全國（文））設(shè)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】由題意結(jié)合復(fù)數(shù)的運(yùn)算法則即可求得z的值.【詳解】由題意可得：SKIPIF1<0.故選：C.10．（2021·全國（理））設(shè)SKIPIF1<0，則SKIP