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已知复数z=【(-1+3i)*(1-i)-(1+3i)】/i,w=z+ai,当|w/z|≤根号2时,求a的取值范围
人气:233 ℃ 时间:2020-06-24 13:13:52
解答
(1-√2)(1+i) < a < (√2+1)(1+i)要过程啊------->z=(1+i)/i
--------->w= (1+i)/i + ai
------------> | w/z | = 1+ai/(1+i)*i<= √2
-----------------> -√2<= w/z <=√2
--------------------->√2<= 1-a/(1+i)<=√2

....... > ( 1-√2)(1+i)
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