Find the solutions of
$\sin z = 3$
There are 2 ways to solve this, I know how to do this with:
$\sin z = \frac{1}{2i}(e^{iz}-e^{-iz}) = 3$
Now, I am now doing in the way:
$\sin z = \sin x \cosh y+i \cos x \sinh y$ = 3
By comparing the terms:
$\sin x \cosh y = 3$
$\cos x \sinh y = 0$
After this part, I have got no idea what to do.
Could anyone help me please?
Thank you.
Best Answer
you have $$\sin x \cosh y = 3, \cos x \sinh y = 0 $$ take the second equation. you have $$\sinh y = 0 \to y = 0 \\ \cos x = 0, x = \pm \pi/2 +2k\pi $$
putting $y = 0,$ in the first equation gives $\sin x = 3$ which has no real solution. we are now left with $$x = \pm \pi/2 +2k\pi \to \cosh y = \pm 3.$$ since $\cosh y \ge 1,$ we only need to solve $$\cosh y = 3 \to e^{2y} - 6e^y+1=0\to e^y = 3\pm2\sqrt2 \to y = \ln(3\pm2\sqrt2)$$
the solutions are $$\sin^{-1}(3)=\pm \pi/2 +2k\pi+i \ln(3\pm2\sqrt2).$$