6楼: Originally posted by hank612 at 2014-06-19 05:14:53
当x+y,  x*y都是实数时,   (x+y)^2= |x|^2 +|y|^2 + \overline{x}y  + x\overline{y}
且  f(x,y)=2xy- \overline{x}y  - x\overline{y} =0 \mathrm{\ or\ } 2r^2(1-\cos(2\theta)) \geq 0

因此, 当x+y,  x*y都 ...

2楼: Originally posted by shuxue0 at 2014-06-16 22:54:43
Consider the two roots of (1.5), denoted by e1 and e2.

The condition |\lambda|<1 is equivalent to
|e1e2|<1
and
|e1+e2|<2
(Notice that e1 and e2 are conjugate if they are not real num ...

3楼: Originally posted by felix2018 at 2014-06-16 23:12:09
很不错的想法，如果一根模大于一，另一根模小于1呢？谢谢你的回复1...