Find all solutions to $\sin(z+i)=1$ So basically for this question I tried going about it two different ways. The first way I used the identity:
$$\sin(z)=\frac{(e)^{iz}-e^{-iz}} {2i}$$
and applying it to z+i as in a translation to +i.
I got $$\sin(z+i)=\frac{(e)^{i(z+i)}-e^{-i(z+i)}} {2i}$$
$$=\frac{(e)^{iz-1}-e^{-iz+1}} {2i}$$
I'm not sure if this is the correct approach or not however. As I also tried this second approach with the basic formula of (knowing $z=x+iy$):
$\sin(z)=\sin(x+iy)=\sin(x)\cos(iy)+\cos(x)\sin(iy)$
$=\sin(x)\cosh(y)+i\cos(x)\sinh(y)$
But then again I need to take into considerations the translation of the $i$. But wouldn't that make my equations into something messy like:
$\sin(z+i)=\sin(x+i(y+1))=\sin(x)\cos(i(y+1))+\cos(x)\sin(i(y+1))$
Is there an identity I'm missing here? I believe this second method I have laid out is the better way to go about this one. But I am not sure.
 A: The simpler solution is $$ z+i=\frac{\pi}{2}+2k\pi$$
$$
z=\frac{\pi}{2}+2k\pi-i
$$
this comes from:
$$
\frac{e^{ix}-e^{-ix}}{2i}=1
$$
that, with $u=e^{ix}$, becomes:
$u^2-2iu-1=0$
$(u-i)^2=0$
$u=i$
so we have
$e^{ix}=i $ and $x=\frac{\pi}{2}+2k\pi$
A: You found $\dfrac{(e)^{iz-1}-e^{-iz+1}} {2i}=1$ with $e^{iz}=w$ then $\dfrac{w}{e}-\dfrac{e}{w}=2i$ or $w^2-2iew-e^2=0$ and $(w-ie)^2=0$ means $e^{iz}=ie$ and
$$iz=\ln|ie|+i(2k\pi+\dfrac{\pi}{2})$$
so
$$\color{blue}{z=-i+2k\pi+\dfrac{\pi}{2}}$$
A: Let $z= x+iy$. Then,
$$\sin (z+1)= \frac{e^{iz-1} - e^{-iz+1}}{2i} = 1 \Longrightarrow e^{iz-1} - e^{-iz+1} = 2i$$
$$e^{iz-1} - e^{-iz+1}= e^{-(y+1)+xi} - e^{(y+1)-xi}= e^{-(y+1)} (\cos (x) + i \sin (x)) - e^{(y+1)} (\cos (-x)+i \sin (-x) ) = e^{-(y+1)} (\cos (x) + i \sin (x)) - e^{(y+1)} (\cos(x)- i \sin (x) )= 2i \Longrightarrow \cos (x) (e^{-(y+1)}-e^{y+1}) =0$$
Separate in cases. First, suppose $e^{-(y+1)}-e^{y+1}=0$.
$$e^{-(y+1)}-e^{y+1}=0 \Longrightarrow -(y+1)= y+1 \Longrightarrow y= -1$$
$$\sin (x) (e^{-(y+1)}+e^{y+1} )= 2 \Longrightarrow \sin(x) = \frac{2}{e^{-(y+1)}+e^{y+1}}= \frac{2}{1+1}=1 \Longrightarrow x = 2 \pi k + \frac{\pi}{2}$$
We then get the first solution: $z= (2 \pi k+ \frac{\pi}{2})-i$.
Now, suppose it was $\cos (x) =0$. Then $sin(x)=1 \lor \sin(x)=-1$. Suppose $\sin (x)= 1$. Then, $x= \frac{\pi}{2}+ 2 \pi k$.
$$\sin (x) (e^{-(y+1)}+e^{y+1} )= 2 \Longrightarrow (e ^ {y+1})^2-2(e^{y+1})+1= (e^{y+1}-1)^2=0 \Longrightarrow y=-1$$  
This solution was included in the case before.
Suppose $\cos (x) = 0$ and $\sin (x) = -1 $. Then, $x=\frac{3 \pi}{2}+ 2 \pi k$.
$$\sin (x) (e^{-(y+1)}+e^{y+1} )= 2 \Longrightarrow e^{-(y+1)}+e^{y+1} = -2$$
Which isn't possible because the exponential function is always positive (for real exponent).
Then, the only solutions are of the form $z= (2 \pi k+ \frac{\pi}{2})-i$, with $k \in \mathbb{Z}$.
