Câu 1.
`\quad {2+i}/{1+i}={(2+i)(1-i)}/{(1+i)(1-i)}`
`={2-2i+i-i^2}/{1-i^2}`
`={3-i}/{1+1}=3/ 2 -1/ 2 i `
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Câu 2.
`\qquad z(2-i)=i`
`<=>z=i/{2-i}={i(2+i)}/{(2-i)(2+i)}`
`<=>z={2i+i^2}/{4-i^2}={2i-1}/{4+1}`
`<=>z={-1+2i}/5=-1/ 5 + 2/ 5 i`
Vậy `z=-1/ 5 + 2/ 5 i`
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Câu 3.
`z=2+i`
`=>\overline{z}=2-i`
`=>z.\overline{z}=(2+i)(2-i)=4-i^2=4+1=5`
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`|\overline{z}|=\sqrt{2^2+(-1)^2}=\sqrt{5}`
Vậy `z.\overline{z}=5` và `|\overline{z}|=\sqrt{5}`
