已知z w 为复数,(1+3i )*z 为纯虚数,w =z /(2+i ),且w 的绝对值=5倍的跟号2,求w各位大哥大姐帮帮小弟.
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![已知z w 为复数,(1+3i )*z 为纯虚数,w =z /(2+i ),且w 的绝对值=5倍的跟号2,求w各位大哥大姐帮帮小弟.](/uploads/image/z/1750395-3-5.jpg?t=%E5%B7%B2%E7%9F%A5z+w+%E4%B8%BA%E5%A4%8D%E6%95%B0%2C%EF%BC%881%EF%BC%8B3i+%EF%BC%89%EF%BC%8Az+%E4%B8%BA%E7%BA%AF%E8%99%9A%E6%95%B0%2Cw+%EF%BC%9Dz+%2F%EF%BC%882%EF%BC%8Bi+%EF%BC%89%2C%E4%B8%94w+%E7%9A%84%E7%BB%9D%E5%AF%B9%E5%80%BC%EF%BC%9D5%E5%80%8D%E7%9A%84%E8%B7%9F%E5%8F%B72%2C%E6%B1%82w%E5%90%84%E4%BD%8D%E5%A4%A7%E5%93%A5%E5%A4%A7%E5%A7%90%E5%B8%AE%E5%B8%AE%E5%B0%8F%E5%BC%9F.)
已知z w 为复数,(1+3i )*z 为纯虚数,w =z /(2+i ),且w 的绝对值=5倍的跟号2,求w各位大哥大姐帮帮小弟.
已知z w 为复数,(1+3i )*z 为纯虚数,w =z /(2+i ),且w 的绝对值=5倍的跟号2,求w
各位大哥大姐帮帮小弟.
已知z w 为复数,(1+3i )*z 为纯虚数,w =z /(2+i ),且w 的绝对值=5倍的跟号2,求w各位大哥大姐帮帮小弟.
设(1+3i)z=bi,则z=(bi)/(1+3i),w=z/(2+i)=(bi)/[(1+3i)(2+i)],|w|=|(bi)/[(1+3i)(2+i)]|=|b|/[√10×√5]=5√2,解得|b|=50,则:w=(bi)/[(1+3i)(2+i)]=±(1-i)
设z=a+bi ,(1+3i )*z 为纯虚数,就是a-3b+bi+3ai,a-3b=0 b+3a不等于0,a=3b 带入z=3b+bi,w =z /(2+i )=(3b+bi)/(2+i )=b(3+i)(2-i )/(2+i )(2-i )=b(7-i)/5
w 的绝对值=5倍的跟号2带入求出b 就可以算出W
设z=a+bi
(1+3i)(a+bi)=(a-3b)+(3a+b)i 是纯虚数
则a-3b=0 a=3b
w=z/(2+i)
=(a+bi)/(2+i)
=(3b+bi)/(2+i)
=b(3+i)(2-i)/[(2+i)(2-i)]
=(b/5)(7-i)
IwI=√[(b/5)^2(49+1)]=5√2
解得b=±5
所以w=(±5/5)(7-i)=7-i或-7+i
希望能帮到你,祝学习进步O(∩_∩)O
答案应该是:7-i 或 i-7