3.2《复数的四则运算》习题(最新整理)

1、3-2-1数系的扩充与复数的引入习题第1课时复数加、减法与乘法的运算法则双基达标 (限时15分钟(1若z132i，z213i，则z12z2.答案18i 2(64i)(64i). 答案523. 如果复数(m2i)(1mi)是实数，则实数m.解析(m2i)(1mi)(m2m)(1m3)ir1m30m1. 答案14. 已知复数z112i，z2m(m1)i，若z1z2的实部与虚部相等，则实数m.解析z1z2(12i)m(m1)im(m1)i2mi2(m1)(2m)(3m1)i，2m3m1m3 ， .43答案45z 3已知 ， 3 3b(b2)i(a，br)若z z 43，则ab 1a(a2 .1)i

2、z212a3 解析zz3 3b(ab1)i4 3，122error!a2，b1，ab3. 答案36. 计算：(1)(2 3i)(3 2)(3 2i)(2i3)；(2)(12i)(2i)(34i)；解(1)原式(2 32 3)(3 3 22)i22i.(2)原式(22i24ii)(34i)(43i)(34i)1212i29i16i25i.综合提高 (限时30分钟(7. 复数(3i1)i的共轭复数是 解析(3i1)i3i，则共轭复数为3i.答案3i8. 设复数z12i，则z22z.解析z22z(z1)21(答案32i)213.9. 若x是纯虚数，y是实数，且2x1iy(3y)i，则xy等于解析由于

3、x是纯虚数，可设xbi(br，b0)，将其代入2x1iy(3y)i得1(2b1)iy(3y)i，error!解得error!xy15 i.215答案 i210已知复数z满足z(12i)103i，则z. 解析设zabi，(a，br)则abi12i103i， 即error!a9，b5.z95i. 答案95i11已知z1(3xy)(y4x)i，z2(4y2x)(5x3y)i(x，yr)设zz1z2且z13 2i，求z1，z2.解zz1z2(3xy)(y4x)i(4y2x)(5x3y)i(3xy)(4y2x)(y4x)(5x3y)i(5x3y)(x4y)i，z(5x3y)(x4y)i. 又z132i，

4、error!解得error!z1(321)(142)i59i，z24(1)22523(1)i87i.12z1iz2azb1iab已知 ，z2z1 ，求实数 ，的值解z1i，z22i，z2z1i，z2azb(ab)(a2)i，z2azb(1i)i1i，(ab)(a2)i1i，error!解得error!13(创新拓展)已知复数z1i，求实数a，b使az2b解z1i，z(a2z)2.az2b z(a2b)(a2b)i，(a2z)2(a2)244(a2)i(a24a)4(a2)i.a，b都是实数，由az2b 得error!z(a2z)2，两式相加，整理得a26a80，解得a12，a24，对应得b11

5、，b22.所求实数为a2，b1或a4，b2.“”“”at the end, xiao bian gives you a passage. minand once said, people who learn to learn are very happy people. in every wonderful life, learning is an eternal theme. as a professional clerical and teaching position, i understand the importance of continuous learning, life is

6、 diligent, nothing can be gained, only continuous learning can achieve better self. only by constantly learning and mastering the latest relevant knowledge, can employees from all walks of life keep up with the pace of enterprise development and innovate to meet the needs of the market. this document is also edited by my studio pro