- #1
Craptola
- 14
- 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.
Show that [itex]\arcsin(2) = \frac{\pi}{2}-i\ln (2\pm \sqrt3)[/itex]
I'm fairly certain the way to solve this is to use
[tex]\sin(z)=\frac{1}{2i}(e^{iz}- e^{-iz})[/tex]
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.
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.