Contradiction in solving sin &#x2061;<!-- ⁡ --> ( z ) = 0 ? sin &#x2061;<!-- ⁡

Contradiction in solving $\mathrm{sin}\left(z\right)=0$?
$\mathrm{sin}\left(z\right)=0$
${e}^{-i\left(x+iy\right)}={e}^{i\left(x+iy\right)}$
${e}^{-ix+y}={e}^{ix-y}$
${e}^{ix-y+ix-y}=1$
${e}^{2ix-2y}={e}^{0}$
$⇒2ix-2y=0$
ix=y
z=x+xi
but, $2i\left(x+iy\right)=2iz=0$
$⇒z=0,z=2\pi n$
※ ?
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esperoanow
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 $z=\pi n$

grenivkah3z
You have made a mistake in that
$2ix-2y=0$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=-iy\ne -y$