新高考數(shù)學一輪復習講與練第10講 復數(shù)（練）（解析版）

第02講復數(shù)1．若復數(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)復數(shù)的四則運算，先求出復數(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】由復數(shù)的除法法則求解．【詳解】由SKIPIF1<0，得SKIPIF1<0.故選：C.3．若復數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0B．復數(shù)SKIPIF1<0在復平面上對應的點在第二象限C．復數(shù)SKIPIF1<0的實部與虛部之積為SKIPIF1<0D．SKIPIF1<0【答案】A【分析】根據(jù)復數(shù)的運算法則，化簡得到SKIPIF1<0，結(jié)合復數(shù)的基本概念，共軛復數(shù)的概念，以及復數(shù)的模的計算公式，逐項判定，即可求解.【詳解】由題意，復數(shù)SKIPIF1<0，可得SKIPIF1<0，所以A正確；復數(shù)SKIPIF1<0在復平面對應的點SKIPIF1<0位于第三象限，所以B錯誤；復數(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；由復數(shù)SKIPIF1<0的共軛復數(shù)為SKIPIF1<0，所以D錯誤.故選：A.4．若復數(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)共軛復數(shù)的定義，求出SKIPIF1<0，再將SKIPIF1<0轉(zhuǎn)化為復數(shù)的標準形式即可.【詳解】由題意，SKIPIF1<0，SKIPIF1<0，∴其虛部為SKIPIF1<0；故選：D.5．已知復數(shù)SKIPIF1<0（SKIPIF1<0為虛數(shù)單位）在復平面內(nèi)對應的點在第三象限，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】利用復數(shù)代數(shù)形式的乘除運算化簡，再由實部與虛部均小于0列不等式組求解.【詳解】因為SKIPIF1<0，在復平面內(nèi)對應的點在第三象限，SKIPIF1<0，解得SKIPIF1<0.故選：A.6．已知復數(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)復數(shù)的運算法則，求得SKIPIF1<0，結(jié)合模的計算公式，即可求解.【詳解】由題意，復數(shù)SKIPIF1<0，所以SKIPIF1<0.故選：B.二、填空題7．已知復數(shù)SKIPIF1<0SKIPIF1<0為純虛數(shù)，則SKIPIF1<0______.【答案】4【分析】由復數(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)定運算SKIPIF1<0，若SKIPIF1<0，設SKIPIF1<0為虛數(shù)單位，則復數(shù)SKIPIF1<0__________.【答案】SKIPIF1<0【分析】根據(jù)新定義運算直接列方程求解.【詳解】因為規(guī)定運算SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，故答案為：SKIPIF1<09．若復數(shù)SKIPIF1<0為實數(shù)，則實數(shù)SKIPIF1<0________．【答案】SKIPIF1<0【分析】根據(jù)復數(shù)代數(shù)形式的乘方及加法運算化簡，再根據(jù)復數(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SKIPIF1<0，由復數(shù)相等，得出SKIPIF1<0的關系式，消去SKIPIF1<0得到關于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)復數(shù)的除法及加法法則，結(jié)合復數(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SKIPIF1<0，代入原方程，根據(jù)復數(shù)相等和SKIPIF1<0可得答案.【詳解】設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SKIPIF1<0,由SKIPIF1<0列方程，即可求出b，進而得到復數(shù)z.【詳解】由題意可設：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的共軛復數(shù)為SKIPIF1<0C．SKIPIF1<0對應的點在第二象限 D．SKIPIF1<0【答案】C【分析】根據(jù)已知條件及復數(shù)的除法法則，再利用復數(shù)的概念及共軛復數(shù)，結(jié)合復數(shù)的幾何意義及復數(shù)的摸公式即可求解.【詳解】由SKIPIF1<0，得SKIPIF1<0，對于A，復數(shù)SKIPIF1<0的虛部為SKIPIF1<0，故A不正確；對于B，復數(shù)SKIPIF1<0的共軛復數(shù)為SKIPIF1<0，故B不正確；對于C，復數(shù)SKIPIF1<0對應的點為SKIPIF1<0，所以復數(shù)SKIPIF1<0對應的點在第二象限，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：在復平面內(nèi)，復數(shù)SKIPIF1<0對應的點位于第二象限.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】由復數(shù)的除法和乘法運算化簡復數(shù)，再由復數(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，在復平面內(nèi)對應的點的坐標為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．已知復數(shù)SKIPIF1<0，若SKIPIF1<0，則當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數(shù)SKIPIF1<0，從而可得SKIPIF1<0，復數(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】利用復數(shù)四則運算法則，計算SKIPIF1<0，然后利用復數(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)復數(shù)的幾何意義可得復數(shù)SKIPIF1<0對應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)復數(shù)模的幾何意義判斷復數(shù)z對應點的軌跡，數(shù)形結(jié)合法判斷SKIPIF1<0的范圍.【詳解】由SKIPIF1<0，則z在復平面內(nèi)對應的點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)復數(shù)代數(shù)形式的運算法則以及復數(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)復數(shù)代數(shù)形式的運算法則，共軛復數(shù)的概念以及復數(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】由共軛復數(shù)的概念及復數(shù)的運算即可得解.【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】利用復數(shù)的除法可求SKIPIF1<0，從而可求SKIPIF1<0.【詳解】由題設有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】利用復數(shù)的乘法可求SKIPIF1<0.【詳解】SKIPIF1<0，故選：D.7．（2022·北京）若復數(shù)z滿足SKIPIF1<0，則SKIPIF1<0（

）A．1 B．5 C．7 D．25【答案】B【分析】利用復數(shù)四則運算，先求出SKIPIF1<0，再計算復數(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)合復數(shù)的運算法則即可求得z的值.【詳解】由題意可得：SKIPIF1<0.故選：C.10．（2021·全國（理））設SKIPIF1<0，則SKIP