人教A版選修1-2第三章3.2.2復(fù)數(shù)代數(shù)形式的乘除運算(1)達標(biāo)過關(guān)訓(xùn)練

1、3.2.2復(fù)數(shù)代數(shù)形式的乘除運算(1)一、選擇題1(2019吉林五十五中月考)i為虛數(shù)單位，復(fù)數(shù)eq f(53i,4i)()A1i B1iC1i D1i解析：eq f(53i,4i)eq f(53i4i,4i4i)eq f(2017i3,17)1i，故選C答案：C2(2019山東實驗中學(xué)質(zhì)量檢測)已知復(fù)數(shù)z滿足zi32i，則復(fù)數(shù)z的虛部為()A3i B3iC3 D3解析：zi32i，zeq f(32i,i)eq f(32ii,i2)(3i2i2)23i，復(fù)數(shù)z的虛部為3.故選D答案：D3(2019安徽慧德月考)復(fù)數(shù)z滿足z(1eq r(3)i)|1eq r(3)i|，則z等于()A1eq r(

2、3)i B1Ceq f(1,2)eq f(r(3),2)i Deq f(r(3),2)eq f(1,2)i解析：z(1eq r(3)i)|1eq r(3)i|，zeq f(|1r(3)i|,1r(3)i)eq f(21r(3)i,1r(3)i1r(3)i)eq f(1,2)eq f(r(3),2)i，故選C答案：C4i是虛數(shù)單位，復(fù)數(shù)eq f(7i,34i)()A1i B1iCeq f(17,25)eq f(31,25)i Deq f(17,7)eq f(25,7)i解析：eq f(7i,34i)eq f(7i34i,34i34i)eq f(213i28i4,25)1i.答案：A5已知復(fù)數(shù)ze

3、q f(r(3)i,1r(3)i)(i為虛數(shù)單位)，則z的虛部為()Aeq f(3,4) Beq f(3,4)iCeq f(r(3),4) Deq f(r(3),4)i解析：zeq f(r(3)i,1r(3)i)eq f(r(3)i1r(3)i,1r(3)i1r(3)i)eq f(r(3)i3,4)eq f(3,4)eq f(r(3),4)i，z的虛部為eq f(r(3),4).故選C答案：C二、填空題6已知i是虛數(shù)單位，計算eq f(1i,1i2)_.解析：eq f(1i,1i2)eq f(1i,2i)eq f(1ii,2ii)eq f(i1,2)eq f(1,2)eq f(1,2)i.答案

4、：eq f(1,2)eq f(1,2)i7.eq f(1i43i,2i1i)_.解析：原式eq f(43i4i3,22ii1)eq f(17i,13i)eq f(17i13i,13i13i)eq f(13i7i21,10)2i.答案：2i8設(shè)a，bR，abieq f(117i,12i)(i為虛數(shù)單位)，則ab的值為_解析：由題意得(abi)(12i)117i，即a2aibi2b117i，a2b(b2a)i117i.eq blcrc (avs4alco1(a2b11，,b2a7，)解得eq blcrc (avs4alco1(a5，,b3.)ab8.答案：8三、解答題9(2019遼寧實驗中學(xué)月考)

5、已知z11i，z222i.(1)求z1z2；(2)若eq f(1,z)eq f(1,z1)eq f(1,z2)，求z.解：(1)z1z2(1i)(22i)4.(2)由eq f(1,z)eq f(1,z1)eq f(1,z2)，得zeq f(z1z2,z1z2)，zeq f(4,1i22i)eq f(4,3i)eq f(62i,5).10已知復(fù)數(shù)z3bi(bR)，且(13i)z為純虛數(shù)(1)求復(fù)數(shù)z；(2)若eq f(z,2i)，求復(fù)數(shù)的模|.解：(1)(13i)z(13i)(3bi)(33b)(9b)i，(13i)z是純虛數(shù)，33b0且9b0，即b1.z3i.(2)eq f(z,2i)eq f(3i,2i)eq f(3i2i,2i2i)eq f(7i,5)eq f(7