專(zhuān)題十三 數(shù)系的擴(kuò)充與復(fù)數(shù)的引入第三十三講 復(fù)數(shù)的計(jì)算答案

1、專(zhuān)題十三數(shù)系的擴(kuò)充與復(fù)數(shù)的引入第三十三講復(fù)數(shù)的計(jì)算答案部分1D【解析】，其共軛復(fù)數(shù)為，對(duì)應(yīng)的點(diǎn)為11i1i11i1i(1i)(1i)222?????????11i22?，故選D11()22?2C【解析】因?yàn)?，所以，故選C21i(1i)2i=2ii2ii1i(1i)(1i)????????????z|z|1?3D【解析】，故選D??i23i32i????4D【解析】故選D2(1i)(2i)2i2ii3i????????5B【解析】因?yàn)椋?/p>

2、以復(fù)數(shù)的共軛復(fù)數(shù)為故選B22(1i)1i1i(1i)(1i)???????21i?1i?6C【解析】由為純虛數(shù)知選C2(1)2ii??7B【解析】由復(fù)數(shù)的運(yùn)算法則，，選B2(1i)(2i)123ii13i????????8C【解析】∵，∴復(fù)數(shù)在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)，位于i(2i)12iz??????z(12)Z??第三象限，選C9A【解析】由，得，，選Ai1iz??1i1iiz????22(1i)2iz????10B【解析】，因?yàn)閷?duì)應(yīng)的點(diǎn)在

3、第二象限，∴(1i)(i)(1)(1)izaaa???????1010aa???????，解得1a??，故選B.11A【解析】因?yàn)?，由已知的，(12i)(i)a??(2)(21)iaa???221aa???解得故選A3a??12C【解析】由得，，所以，故選Ci3iz???32zi??32zi??13D【解析】，故選D224343||5543ziiz?????D32A【解析】??????2122211112iiiiziiii??????

4、??????33D【解析】????325zi???，得53552zizii???????34A【解析】設(shè)zabi??，則zabi??，由22zziz???得，??????222222abiabiiabiabi????????izbaa????????????????111222bba22所以選A35C【解析】2442izii????對(duì)應(yīng)的點(diǎn)的坐標(biāo)是??42?故選C36C【解析】由知，，所以4MN??4zi?4zi??37D【解析】211

5、izii????，1zi???。38A【解析】，選A??212iii???39B【解析】設(shè)表示復(fù)數(shù)，則的共軛復(fù)數(shù)對(duì)應(yīng)的點(diǎn)位()Axyzxyi??zzxyi??()Bxy?40B【解析】由已知，所以111(1)(1)22iziii??????????2||2z?41D【解析】∵==，∴的共軛復(fù)數(shù)為，故選D.z32ii???1i??z1i??42A【解析】由對(duì)應(yīng)復(fù)平面內(nèi)的點(diǎn)為A1010(3)133(3)(3)iiiiiii???????43