Finding arcsin(2) using complex exponentials

  • Thread starter Craptola
  • Start date
2 replies · 2K views · last reply
Craptola
Messages
14
Reaction score
0
Got a maths exam tomorrow been looking through some past papers. Have hit a stumbling block with regard to complex numbers, the problem lies with my algebra.

Homework Statement


Show that [itex]\arcsin(2) = \frac{\pi}{2}-i\ln (2\pm \sqrt3)[/itex]


Homework Equations


I'm fairly certain the way to solve this is to use
[tex]\sin(z)=\frac{1}{2i}(e^{iz}- e^{-iz})[/tex]


The Attempt at a Solution


Equating sin(z) to 2 I could only rearrange it to

[tex]4i=e^{iz}-e^{-iz}[/tex]
I was always pretty awful at algebra and can't see a way to rearrange for z. If anyone could nudge me in the right direction it would be greatly appreciated.
 
The problem becomes simpler if you make the substitution [itex]v=e^{iz}[/itex].