e^(iz)=cos(z)+isin(z), e^(-iz)=cos(z)-isin(z).
Therefore, e^(iz)-e^(-iz)=2isin(z),
i(e^(iz)-e^(-iz))=-2sin(z),
sin(z)=-(i/2)(e^(iz)-e^(-iz)).
f(z)=sin(z)=0⇒iz=-iz, 2iz=0⇒z=0, implying that the real and imaginary components are both zero.
sin(z)=0 when z=nπ (no imaginary part).
There are no complex zeroes for f(x)=sin(z).