I made a general formula for $sin(z)=R$, and after looking at the imaginary part, I found that it exactly equaled $-arccosh(R)$. I was wondering if I did this correctly and if there is anything related:
- General formula for $sin(z)=R$ (Based on a blackpenredpen video about sin(z)=2):
$sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$
$e^{iz}-e^{-iz}=2iR$
$e^{iz}$ → $e^{2iz}-1=2iRe^{iz}$
$e^{2iz}-2iRe^{iz}-1=0$
$e^{iz}=iR\pm\sqrt{1-R^{2}}$
$iz=\ln\left(iR\pm\sqrt{1-R^{2}}\right)$
$iz=\ln\left(i\left(R\pm\sqrt{R^{2}-1}\right)\right)$
$iz=\ln\left(i\right)+\ln\left(R\pm\sqrt{R^{2}-1}\right)$
$iz=i\frac{\pi}{2}+\ln\left(R\pm\sqrt{R^{2}-1}\right)$
$z=\frac{\pi}{2}-i\ln\left(R\pm\sqrt{R^{2}-1}\right)$
I know I left out a +$2\pi n$, also this can be verified by plugging back into the complex sine formula
- Using this equation to get to my equation
$z=\frac{\pi}{2}-i\ln\left(R\pm\sqrt{R^{2}-1}\right)$
$\Im (z)=-\ln\left(R\pm\sqrt{R^{2}-1}\right)$
After looking at this function in Desmos I realized that this is equal to $-\pm\ arcosh(R)$
This is pretty weird and interesting so I was wondering if anyone could check my work to see if I'm right. I was also wondering if this is useful at all, or if there is anything I could do with this.
Thanks!