--------->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)
推荐
- 已知复数z=(-1+3i)*(1-i)/i-(1+3i)/i,w=z+ai(a属于R),当|w/z|≤根号2时,求a的取值范围
- 已知复数z=(-1+3i)(1-i)/i-(1+3i)/i,w=z+ai,a属于R.当Iw/zI小于等于根号2,求a的取值范围
- 已知复数z=(-1+3i)*(1-i)-(1+3i),w=z=ai(a属于R),当|w/z|≤根号2时,求a的取值范围.
- 已知复数Z满足(1+i)Z=1+根号3i,则|Z|=
- 已知z,ω为复数,(1+3i)•z为纯虚数,ω=z/2+i,且|ω|=52,求复数z及ω(设z=x+yi,x、y∈R)
- wherecanborrowbooksfromourschooll_____.
- 证明:函数f(x)=x+1/x在(0,1)上为减函数.
- in which i spent most of the time at the age of one翻译中文谢谢!
猜你喜欢