Complex Arcsin possible definition

Question

Is this a valid definition for arcsin? $$\sin\theta=z$$ $$e^{i\theta}=\cos\theta+i\sin\theta$$ $$-\biggl(e^{-i\theta}=\cos\theta-i\sin\theta\biggr)$$ $$e^{i\theta}-e^{-i\theta}=2i\sin\theta$$ $$\frac{e^{i\theta}-e^{-i\theta}}{2i}=\sin\theta$$ $$\frac{ie^{-i\theta}-ie^{i\theta}}{2}=\sin\theta$$ $$\frac{ie^{-i\theta}-ie^{i\theta}}{2}=z$$ $$\frac{i}{e^{i\theta}}-ie^{i\theta}=2z$$ $$\frac{-1}{ie^{i\theta}}-ie^{i\theta}=2z$$ $$\frac{-1-ie^{2i\theta}}{ie^{i\theta}}=2z$$ $$-1-ie^{2i\theta}=2zie^{i\theta}$$ $$ie^{2i\theta}+2zie^{i\theta}+1=0$$ $$(ie^{i\theta}+z)^2-z^2+1=0$$ $$(ie^{i\theta}+z)^2=z^2-1$$ $$ie^{i\theta}+z=\pm\sqrt{z^2-1}$$ $$ie^{i\theta}=-z\pm\sqrt{z^2-1}$$ $$e^{i\theta}=(z\mp\sqrt{z^2-1})i$$ $$e^{i\theta}=(z\mp\sqrt{z^2-1})e^{\frac{i\pi}{2}}$$ $$i\theta=\ln(z\mp\sqrt{z^2-1})+\frac{i\pi}{2}$$ $$\theta=\frac{\pi}{2}-\ln(z\mp\sqrt{z^2-1})i$$ Therefore as $\arcsin(z)=\theta$ $$\arcsin(z)=\frac{\pi}{2}-\ln\Bigl(z\mp\sqrt{z^2-1}\Bigr)i = \theta +2A\pi$$ you can do ln of any complex number so this should be defined for all complex numbers right? Unless its incorrect