Proof required for expressing ARCSIN in complex exponential function.

**jalal0** · April 10th 2008, 10:09 AM

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)]

Thanks in advance.

**bobak** · April 10th 2008, 10:34 AM

Start with

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

$(1) - (2)$

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

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

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

you should get

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

use the quadratic formula.

to get $e^{iz} = i \sin z \pm \sqrt{1 - \sin^2 z}$

then take logs $z = -i \ \ln (\ i \sin z \pm \sqrt{1 - \sin^2 z} \ )$

remember $i^2 = -1$ and $\frac{1}{i} = -i$

Bobak

**jalal0** · April 10th 2008, 10:32 PM

Thanks a lot mate. :-)

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Proof required for expressing ARCSIN in complex exponential function.

thanks

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