The correct options are
A √3i
B 0
C −√3i
z2−z+1=0⇒z=1±i√32
⇒z=−ω or z=−ω2
If z=−ω, then
zn−z−n=(−ω)n−(−ω)−n
=(−1)nωn−(−1)−nωn
=ωn−ω2n
Here, there are two cases possible
Case 1:n is multiple of 3
=ωn−ω2n=1−1=0
Case 2:n=3m+1,m∈Z+
=ω3m+1−ω2(3m+1)=ω3mω−ω6mω2
=ω−ω2=√3i
Case 3:n=3m+2,m∈Z+
=ω3m+2−ω2(3m+2)=ω3mω2−ω6mω4
=ω2−ω=−√3i
If z=−ω2, then
zn−z−n=(−ω2)n−(−ω2)−n
=(−1)nω2n−(−1)−nω2n
=ω2n−ωn
Similarly as above we can do.