# Find $\sin^{-1} (-2)$ on the complex value.

Q: Find the inverse sine $\sin^{-1} (-2)$ by writing $\sin(w) = -2$

ii) Using the Def'N $\sin w = \frac {e^{iw} - e^{-iw}}{2i}$

This was actually a 2 part question the first part was to find the inverse using rectangular representation for $\sin w$ and wasn't too bad.

My attempt: $\frac {e^{iw} - e^{-iw}}{2i} = -2$

$e^{iw} - e^{-iw} = -4i$

$e^{iw} (e^{iw} - e^{-iw}) = e^{iw} (-4i)$

$(e^{2iw} - e^{0}) = -4i e^{iw}$

$(e^{2iw} + 4i e^{iw} - 1 ) = 0$ let $a =e^{iw}$

$(a^2 + 4ia -1)= (a^2 + 4ia -4 +4 -1) = (a+2i)^2 +4 -1 = (a+2i)^2 +3$

$(a+2i) = \pm i \sqrt 3$

$a = -i (2 \pm \sqrt 3 )$

$e^{iw} = -i (2 \pm \sqrt 3 )$

$iw = \ln[ {-i (2 \pm \sqrt 3 )} ]$

$w =\frac {\ln[ {-i (2 \pm \sqrt 3 )} ]}{i}$

Honestly that doesn't look like it makes any sense.

Any ideas how to do this?

Here $$e^{2iw} + 4i e^{iw} - 1 = 0$$ then $e^{iw}=-2i+\sqrt{-3}$ and $$iw=\log[-2i+\sqrt{-3}]=\log[-i(2-\sqrt{3})]=\ln(2-\sqrt{3})+i\arg[-i(2-\sqrt{3})]=\ln(2-\sqrt{3})+i(-\dfrac{\pi}{2}+k\pi)$$ so $$w=-\dfrac{\pi}{2}+k\pi-i\ln(2-\sqrt{3})$$ for main branch $k=0$.
rewrite $−i(2±√3)$ as $e^{ix}$ then $w=x$