solve $\sin(z)=-1$ in the set of complex numbers

Question

I'm pretty sure I'm on the right track, but am I missing anything? Can anything further be done with this?

Solve $\sin(z)=-1$ in the set of complex numbers.

$\sin(z)=-1$

$\Rightarrow{e^{iz}-e^{-iz} \over 2i} =-1$

$\Rightarrow e^{iz}-e^{-iz} =-2i$

$\Rightarrow e^{iz}-{1 \over e^{iz}} + 2i =0$

$\Rightarrow (e^{iz})^2-1 + 2ie^{iz} =0$

for simplicity say $x=e^{iz}$

$\Rightarrow x^2-2xi-1 =0$

$\Rightarrow x=-i$

$\Rightarrow e^{iz}=-i$

$\Rightarrow z={\ln(-i) \over i} $

Note that $$ e^{iz}=-i\implies e^{iz}=e^{-i(\pi/2 -2n\pi)}\implies z=-\frac{\pi}{2}+2n\pi$$ — Mark Viola, May 05 '16 at 04:10
@copper.hat So should I leave my answer as $\Rightarrow e^{iz}=-i$ or is there a way to solve for all values of z that are solutions? — chris, May 05 '16 at 04:22
@Dr.MV Can you please explain why $-i=e^{-i(\pi/2 -2n\pi)}$? — chris, May 05 '16 at 04:24
@Dr.MV After some thinking, I understand why that is true and how it helps. Thanks. — chris, May 05 '16 at 05:02

score 2 · Answer 1 · answered May 05 '16 at 04:47

2

$\sin z=-1\implies\cos z=0\implies e^{iz}=-i=\cos\dfrac\pi2-i\sin\dfrac\pi2=e^{-i\pi/2}$

$$z=2n\pi-\dfrac\pi2$$

answered May 05 '16 at 04:47

lab bhattacharjee

1 Answers1