If `e^(itheta)=costheta+is intheta,` find the value of `|1e^(ipi//3)e^(ipi//4)e^(-ipi//3)1e^(i2pi//3)e^(-ipi//4)e^(-i2pi//3)1|`

Answer

`-2-sqrt(2)`

Explanation

Expanding by Sarrus rule.
`|{:(1,,e^(i pi//3),,e^(i pi//4)),(e^(ipi//3),,1,,e^(i2pi//3)),(e^(-i2pi//3),,e^(-i2pi//3),,1):}|= 1+e^(I pi//3).xx e^(i2pi//3) xx e^(-ipi//4)+e^(-I pi//3)`
`xxe^(-i2pi//3)xxe^(ipi//4)-(e^(-ipi//4)xxe^(-ipi//3)xxe^(-ipi//3)+e^(i2pi//3)xxe^(i2pi//3))`
`=1+e^(i3pi//4)+e^(-i3pi//4) -(1+1+1)`
`=-2+2cos (3pi//4)`
`=-2-sqrt(2)`