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To solve for the $\sin^{-1} z$ ($z$ element of $\mathbb{C}$), the book reads that

  1. $w = \sin^{-1} z$ when $z = \sin w$ implies:

  2. $w = \sin^{-1} z$ when $z = (e^{iw} - e^{-iw})/2i$ implies:

  3. $(e^{iw})^2 - 2iz(e^{iw}) - 1 = 0$ *

I don't see (2) implies (3). I mean it looks like the first step they used was mutiply out the $2i$, then subtract $2zi$ from both sides, but I'm lost after that.

2 Answers 2

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If $z = \frac{e^{iw}-e^{-iw}}{2i}$, then $2iz = e^{iw}-e^{-iw}$. Multiplying through by $e^{iw}$ you get $2ize^{iw} = (e^{iw}-e^{-iw})e^{iw} = e^{iw}e^{iw}-e^{-iw}e^{iw} = (e^{iw})^2 - 1.$

(Why would you multiply through by $e^{iw}$? Because $e^{iw} - e^{-iw} = e^{iw} - \frac{1}{e^{iw}} = \frac{e^{iw}e^{iw} - 1}{e^{iw}}$ so you "really" have $2iz = \frac{e^{iw}e^{iw}-1}{e^{iw}}$, so multiplying through by $e^{iw}$ just clears denominators, just like the initial multiplication by $2i$).

So $-2ize^{iw} - 1 = -(e^{iw})^2 + 1 - 1 = -(e^{iw})^2$.

Adding $(e^{iw})^2$ will give you $0$.

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$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$

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    I believe you misread the question. This explains why (1) implies (2) but isn't really relevant to getting from (2) to (3)2011-02-24