Let `z=1+ai` be a complex number, `a > 0`,such that `z^3` is a real number. Then the sum `1+z+z^2+...+ z^11` is equal to:

Options

A. `-1250sqrt(3)i`
B. `1250sqrt(3)i`
C. `-1365sqrt(3)i`
D. `1365sqrt(3)i`

Answer

C

Explanation

It is given that z=1+ai is such that `z^(3)` is real `rArr 3a-a^(3)=0 rArr a=sqrt(3)` `[therefore a lt 0]`
`therefore z=1+ai rArr z=1+isqrt(3)=2(cospi/3+isinpi/3) = 2e^(ipi//3)`
`rArr z^(12)=2^(12)e^(i4pi)=2^(12)`
`therefore 1+z+z^(2)+.....+z^(11)=(1-z^(12))/(1-z)=(1-2^(12))/(1-(1+isqrt(3)))=-4095/-(isqrt(3))`
`=-1365 sqrt(3)i`