Since f(z) is a periodic function we can arbtrarily set f(z)=sin(kz) where k can be complex.
Because the period is i f(z)=f(z+ni)=sin(k(z+ni))=sin(kz+kni), where n is any integer.
Therefore kni=2πn and k=2π/i=-2πi, and f(z)=sin(-2πzi) or -sin(2πzi). The sign is irrelevant because sin(2πzi) is nevertheless periodic.
To confirm this, if f(z)=sin(2πzi), then f(z+ni)=sin(2πi(z+ni))=sin(2πzi-2πn)=sin(2πzi)=f(z).
g(z)=f(iz-2)=sin(2πi(iz-2))=sin(-2πz-4πi)=-sin(2πz+4πi).
If the period for z is P,
g(z+nP)=-sin(2π(z+nP)+4πi)=-sin(2πz+2πnP+4πi).
To be periodic, 2πnP=2πn, so P=1.
The period of g(z) is 1.