$\mathrm{sin}(z)=0$

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

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

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

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

$\Rightarrow 2ix-2y=0$

ix=y

z=x+xi

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

$\Rightarrow z=0,z=2\pi n$

※ ?

hornejada1c
2022-07-08
Answered

Contradiction in solving $\mathrm{sin}(z)=0$?

$\mathrm{sin}(z)=0$

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

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

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

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

$\Rightarrow 2ix-2y=0$

ix=y

z=x+xi

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

$\Rightarrow z=0,z=2\pi n$

※ ?

$\mathrm{sin}(z)=0$

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

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

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

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

$\Rightarrow 2ix-2y=0$

ix=y

z=x+xi

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

$\Rightarrow z=0,z=2\pi n$

※ ?

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