【答案】
分析:(1)由题意可先令虚数z=x+yi(x,y∈R且y≠0),代入
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024183308915695525/SYS201310241833089156955020_DA/0.png)
,整理后令虚部为0,解出x
2+y
2=4(y≠0),即可求得此虚数的模;
(2)由|z-2|=2可得(x-2)
2+y
2=4,与(1)的结论方程x
2+y
2=4(y≠0)联立,解此方程组,即可得到复数z,代入
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024183308915695525/SYS201310241833089156955020_DA/1.png)
即可解出a的值
解答:解:设z=x+yi(x,y∈R且y≠0)(2分)
则
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024183308915695525/SYS201310241833089156955020_DA/2.png)
∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024183308915695525/SYS201310241833089156955020_DA/3.png)
(4分)
∴x
2+y
2=4(y≠0),即|z|=2; (6分)
又|z-2|=2得 (x-2)
2+y
2=4,与x
2+y
2=4(y≠0)联立
解得
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024183308915695525/SYS201310241833089156955020_DA/4.png)
或
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024183308915695525/SYS201310241833089156955020_DA/5.png)
∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024183308915695525/SYS201310241833089156955020_DA/6.png)
(10分)
∴a=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024183308915695525/SYS201310241833089156955020_DA/7.png)
=2 (12分)
点评:本题考查复数代数形式的混合运算,考查复数的乘法,求复数的模,复数求模公式,解题的关键是用待定系数法设出复数的代数形式,以及理解虚数z满足
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024183308915695525/SYS201310241833089156955020_DA/8.png)
(其中a为实数),得出虚部为0,从而得到复数的实部与虚部所满足的方程.本题考查了待定系数法,其特征是所研究的对象性质已知,可根据其性质设出它的解析式.