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) = -ilog[iz ± √(1-z²)]

Thanks in advance.

- Apr 10th 2008, 11:09 AMjalal0Proof 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)]

Thanks in advance. - Apr 10th 2008, 11:34 AMbobak
Start with

giving

rewrite this as a quadratic in

you should get

use the quadratic formula.

to get

then take logs

remember and

Bobak - Apr 10th 2008, 11:32 PMjalal0thanks
Thanks a lot mate. :-)