If `z=(sqrt3/2+i/2)^5+(sqrt3/2-i/2)^5`, then

Options

A. `Re(z) =0`
B. `Im ( z) =0`
C. `Re(z) gt 0 ,Im(z) gt 0 `
D. `Re(z) gt 0,Im (z) lt 0`

Answer

B

Explanation

Given `z=(sqrt(3)/2+i/2)^5+(sqrt(3)/2-i/2)^5`
`[because omega=(-1+isqrt(3))/(2) and omega^2=(-1-isqrt(3))/(2)]`
Now , `sqrt(3+i)/2=-i((-1+isqrt(3))/2)=-iomega`
and `(sqrt(3)-1)/2=((-1-isqrt(3))/2)=i omega^2`
`therefore z=(-iomega)^5+(iomega^2)^5=iw^2+iw`
`=i(omega-omega^2)=i(isqrt(3))=-sqrt(3)`
`rArr Re(z) lt 0 andlm (z)=0`
Alternate Solution
We know that `z+bar z =2 Re (z)`
If ` z=(sqrt(3)/2+i/2)^5 +(sqrt(3)/2-i/2)^5` then z is purely real ,i,e Im (z)=0