Contradiction in solving sin(z)=0?

Question

$sin(z)=0$

$\Rightarrow e^{-iz}-e^{iz}=0$

$e^{-i(x+iy)}=e^{i(x+iy)}$

$e^{-ix+y}=e^{ix-y}$

$e^{ix-y+ix-y}=1$

$e^{2ix-2y}=e^0$

$\Rightarrow 2ix-2y=0$

$ix=y$

$z=x+xi$

but, $2i(x+iy)=2iz=0$

$\Rightarrow z=0, z=2πn$

※ ?

Answer 1

The exponential is no longer injective in the complex plane. When you have $e^{2ix-2y}=1=e^0$ you can't conclude $2ix-2y=0$. What happens is that $1=e^{2\pi in}$, so you have to solve $2ix-2y=2\pi in$ this gives the missing solutions $z=\pi n$

Answer 2

You have made a mistake in that $$2ix-2y=0$$ $$\implies x=-iy\ne-y$$