When considering complex functions I know that say $f(z)=e^z=e^{x+iy}=e^x(\cos(y)+i\sin(y))$. So now consider $f(z)=\cos(z).$
My question is :
Following the same logic used for $e^z$ does this the mean that $\cos(z)=\cos(x+iy)=\cos(x)\cos(iy)-\sin(x)\sin(iy)$ (by trig identities) ?
Yes, that is correct. I mean, that's not the definition, but it is correct nevertheless. Now, you should add to it that$$\cos(iy)=1+\frac{y^2}{2!}+\frac{y^4}{4!}+\cdots=\cosh(y)$$and that$$\sin(iy)=iy+i\frac{y^3}{3!}+i\frac{y^5}{5!}+\cdots=i\sinh(y).$$