# Proof required for expressing ARCSIN in complex exponential function.

• Apr 10th 2008, 10:09 AM
jalal0
Proof required for expressing ARCSIN in complex exponential function.
The question goes this way:

Given the complex inverse sine function w(z)=sin-¹(z) is defined such that z=sin(w).

Writing sin(w) in terms of complex exponential function, show that:

arcsin(z) = - i log[ i z ± √(1-z²)]

• Apr 10th 2008, 10:34 AM
bobak

$\displaystyle e^{iz} = \cos z + i \sin z \ \ \ \ (1)$
$\displaystyle e^{-iz} = \cos z - i \sin z \ \ \ (2)$

$\displaystyle (1) - (2)$

$\displaystyle e^{iz} - e^{-iz} = 2 i \sin z$

giving $\displaystyle \sin z = \frac{ e^{iz} - e^{-iz}}{2i}$

rewrite this as a quadratic in $\displaystyle e^{iz}$

you should get

$\displaystyle (e^{iz})^2 - 2 i \sin z e^{iz} - 1$

to get $\displaystyle e^{iz} = i \sin z \pm \sqrt{1 - \sin^2 z}$
then take logs $\displaystyle z = -i \ \ln (\ i \sin z \pm \sqrt{1 - \sin^2 z} \ )$
remember $\displaystyle i^2 = -1$ and $\displaystyle \frac{1}{i} = -i$