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问题 解答题

知复数z=（1-i）2+3+6i．

（1）求z及|z|；

（2）若z2+az+b=-8+20i，求实数a，b的值．

答案

（1）z=（1-i）2+3+6i=-2i+3+6i=3+4i，

|z|=

32+42
=5；

（2）z2+az+b=（3+4i）2+a（3+4i）+b=（3a+b-7）+（4a+24）i，

所以z2+az+b=-8+20i，即=（3a+b-7）+（4a+24）i=-8+20i，

所以

3a+b-7=-8
4a+24=20
，解得
a=-1
b=2
；

完形填空

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