Đáp án:
`|z_1+z_2|=2\sqrt{2}`
Giải thích các bước giải:
Gọi $z_1=a_1+b_1i; z_2=a_2+b_2i$
`\qquad (a_1;a_2;b_1;b_2\in RR)`
Vì `|z_1|=|z_2|=\sqrt{3}`
`=> \sqrt{a_1^2+b_1^2}=\sqrt{a_2^2+b_2^2}=\sqrt{3}`
`=>a_1^2+b_1^2=a_2^2+b_2^2=3`
$\\$
Vì `|z_1-z_2|=2`
`=>|(a_1+b_1i)-(a_2+b_2)i|=2`
`=>|(a_1-a_2)+(b_1-b_2)i|=2`
`=>\sqrt{(a_1-a_2)^2+(b_1-b_2)^2}=2`
`=>(a_1-a_2)^2+(b_1-b_2)^2=4`
`=>a_1^2+a_2^2-2a_1a_2+b_1^2+b 2^2-2b_1b_2=4`
`=> -2(a_1a_2+b_1b_2)+(a_1^2+b_1^2)+(a_2^2+b_2^2)=4`
`=> -2(a_1a_2+b_1b_2)+3+3=4`
`=>2(a_1a_2+b_1b_2)=2`
$\\$
`\qquad |z_1+z_1|=|a_1+b_1i+a_2+b_2i|`
`=|(a_1+a_2)+(b_1+b_2)i|`
`=\sqrt{(a_1+a_2)^2+(b_1+b_2)^2}`
`=\sqrt{a_1^2+a_2^2+2a_1a_2+b_1^2+b 2^2+2b_1b_2}`
`=\sqrt{2(a_1a_2+b_1b_2)+(a_1^2+b_1^2)+(a_2^2+b_2^2)}`
`=\sqrt{2+3+3}=2\sqrt{2}`
Vậy `|z_1+z_2|=2\sqrt{2}`
