8楼: Originally posted by 307385553 at 2012-05-30 14:16:16
设x=a+bi,然后把x带入x^2=i中,用待定系数法求解。

9楼: Originally posted by 33221100 at 2012-05-30 23:30:43
应该有两个解：
(1+i)/2^(1/2)
(1-i)/2^(1/2)

10楼: Originally posted by 33221100 at 2012-05-30 23:36:48
后一个写错了：
-(1+i)/2^(1/2)

3楼: Originally posted by dgxs at 2012-05-29 19:23:49
+/-sqrt(i)是什么数？实数？虚数？复数？

15楼: Originally posted by kevin123581 at 2012-05-31 10:37:06
根号i记住就行了，记不住就自己算下。现在的出的结果是x=+/-sqrt（i）=a+bi
平方： i=a^2-b^2+2abi
得到a^2-b^2=0,2ab=1.
==>   a=b=+/- sqrt(2)/2
所以x=sqrt(2)/2+sqrt(2)/2*i
或者x=-sqrt(2)/2-sqrt(2) ...

14楼: Originally posted by zhangjob at 2012-05-31 10:02:19
这个问题对没学过复变函数的人来说，，，是很好奇的，哈哈哈，，，辐角公式啊，±(1+i)/√2。。。。。。。。。