2013-06-06 · 知道合伙人教育行家
关注
展开全部
z=(√2-i)/(1+√2i)²
=(√2-i)/(1+2√2i+2i²)
=(√2-i)/(1+2√2i-2)
=(√2-i)/(2√2i-1)
=[(√2-i)(2√2i+1)]/[(2√2i-1)(2√2i+1)]
=(4i+√2-2√2i²-i)/(8i²-1)
=(3√2+3i)/(-9)
=-√2/3-i/3
则|z|=√[(-√2/3)²+(-1/3)²]=√3/3
【中学生数理化】团队为您解答!祝您学习进步
不明白可以追问!
满意请点击下面的【选为满意回答】按钮,O(∩_∩)O谢谢
=(√2-i)/(1+2√2i+2i²)
=(√2-i)/(1+2√2i-2)
=(√2-i)/(2√2i-1)
=[(√2-i)(2√2i+1)]/[(2√2i-1)(2√2i+1)]
=(4i+√2-2√2i²-i)/(8i²-1)
=(3√2+3i)/(-9)
=-√2/3-i/3
则|z|=√[(-√2/3)²+(-1/3)²]=√3/3
【中学生数理化】团队为您解答!祝您学习进步
不明白可以追问!
满意请点击下面的【选为满意回答】按钮,O(∩_∩)O谢谢
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询