If `z=((sqrt(3))/2+i/2)^5+((sqrt(3))/2-i/2)^5` , then prove that `I m(z)=0.`

Options

A. `R(z) gt 0 and I(z) gt 0`
B. `I(z)=0`
C. `R(z) lt 0 and I(z) gt 0`
D. `R(z)-3`

Answer

B

Explanation

Given `z= (sqrt(3)/2+ i/ 2)^5+(sqrt(3)/2-i/2)^5`
`because ` Euler's form of
`(sqrt(3)/2+i/2=( cos""pi/6+ I sin pi/6 ) = e^i(pi//6)`
`sqrt(3)/2- i/2= cos ((-pi )/(6)+isin ""pi/6)=e^(i(pi//6))`
So, `z= (e^(i pi//6))^5+(e^(-ipi//6))^5=e^(i(5pi)/6)+e^(-i(5pi)/6)`
`=(cos""(5pi)/6+ i sin ""(5pi)/6)+(cos ""(5pi)/6- isin "" (5pi)/6)`
`= 2 cos""(5 pi)/6 " "[ because e^(i theta) = cos theta + i sin theta]`