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.
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.
Start with
$\displaystyle e^{iz} = \cos z + i \sin z \ \ \ \ (1)$
$\displaystyle e^{-iz} = \cos z - i \sin z \ \ \ (2)$
$\displaystyle (1) - (2)$
$\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 $
use the quadratic formula.
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 $\displaystyle \frac{1}{i} = -i$
Bobak