2020-2021學(xué)年高一年級(jí)下冊(cè)學(xué)期數(shù)學(xué)人教B版必修第四冊(cè)第十章課時(shí)作業(yè)10.2.2復(fù)數(shù)的乘法與除法-【含答案】

1、復(fù)數(shù)的乘法與除法一、選擇題1已知復(fù)數(shù)z2i，則zeq o(z,sup6(）的值為()B.eq B.eq r(5)D.eq D.eq r(3)2i是虛數(shù)單位，復(fù)數(shù)eq f(7i,34i)()A1i B1iC.eq f(17,25)eq f(31,25)i Deq f(17,7)eq f(25,7)i3z1，z2是復(fù)數(shù)，且zeq oal(2,1)zeq oal(2,2)0，則正確的是()Azeq oal(2,1)zeq oal(2,2)Bz1，z2中至少有一個(gè)是虛數(shù)Cz1，z2中至少有一個(gè)是實(shí)數(shù)Dz1，z2都不是實(shí)數(shù)4若zeq o(z,sup6(）6，zeq o(z,sup6(）10，則z()A1

2、3i B3iC3i D3i二、填空題5已知a為實(shí)數(shù)，eq f(ai,1i)是純虛數(shù)，則a_.6若(xi)i12i(xR)，則x_.7復(fù)數(shù)eq f(5,2i)的共軛復(fù)數(shù)是_三、解答題8計(jì)算：eq f(32i,23i)eq f(32i,23i).9若z滿(mǎn)足z1eq r(3)(1z)i，求zz2的值10已知復(fù)數(shù)z滿(mǎn)足z(13i)(1i)4.(1)求復(fù)數(shù)z的共軛復(fù)數(shù)；(2)若wzai，且復(fù)數(shù)w對(duì)應(yīng)向量的模不大于復(fù)數(shù)z所對(duì)應(yīng)向量的模，求實(shí)數(shù)a的取值范圍1解析：zeq o(z,sup6(）(2i)(2i)22i2415，故選A.答案：A2解析：eq f(7i,34i)eq f(7i34i,34i34i)e

3、q f(2525i,25)1i，故選A.答案：A3解析：取z112i，z212i滿(mǎn)足zeq oal(2,1)zeq oal(2,2)0，排除A；取z11，z22i滿(mǎn)足zeq oal(2,1)zeq oal(2,2)0，排除D；取z1i，z22i，滿(mǎn)足zeq oal(2,1)zeq oal(2,2)0，排除C，從而選B.答案：B4解析：設(shè)zabi(a，bR)，則eq o(z,sup6(）abi，eq blcrc (avs4alco1(2a6，,a2b210，）解得eq blcrc (avs4alco1(a3，,b1），則z3i.答案：B5解析：eq f(ai,1i)eq f(ai1i,1i1i)

4、eq f(a1a1i,2)，因?yàn)閑q f(ai,1i)是純虛數(shù)，所以a10且a10，即a1.答案：16解析：由題意，得xieq f(12i,i)eq f(i2i2,i2)eq f(i2,1)2i，所以x2.答案：27解析：eq f(5,2i)eq f(52i,2i2i)eq f(52i,5)2i，其共軛復(fù)數(shù)為2i.答案：2i8解：方法一：eq f(32i,23i)eq f(32i,23i)eq f(32i23i32i23i,23i23i)eq f(613i6613i6,49)eq f(26i,13)2i.方法二：eq f(32i,23i)eq f(32i,23i)eq f(i23i,23i)e

5、q f(i23i,23i)ii2i.9解：z1eq r(3)(1z)i，zeq f(1r(3)i,1r(3)i)eq f(1r(3)i2,1r(3)i1r(3)i)eq f(1,2)eq f(r(3),2)i，zz2eq f(1,2)eq f(r(3),2)ieq blc(rc)(avs4alco1(f(1,2)f(r(3),2)i）2eq f(1,2)eq f(r(3),2)ieq blc(rc)(avs4alco1(f(1,2)f(r(3),2)i）1.10解：(1)z1i3i3424i，所以復(fù)數(shù)z的共軛復(fù)數(shù)為24i.(2)w2(4a)i，復(fù)數(shù)w對(duì)應(yīng)的向量為(2,4a)，其模為eq r(44a2)eq r(208aa2).又復(fù)數(shù)z