Question

Does the function f on C defined by f(z) = sin z, z ∈ C, have any complex zeros other than those in the set {nπ : n = 0, ±1, ±2, . . .}. explain

Answer 1

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).