How do I find all solutions to sinh(1/z)=1? I know that $\sinh(z) = (e^z - e^{-z})/2$ but I don't know how to adapt this for the argument $1/z$.
Am I right in assuming $\sinh(1/z) = (e^{1/z} - e^{-1/z})/2?$
 A: How about $z=1/(\operatorname{arcsinh}1+2kπi)$?
A: Let $u + iv = w = 1/z$.  Then solve for $w$ such that $\sinh(w) = 1$, and then solve for $z$.
Note that $\sinh(u + iv) = \sinh(u) \cos(v) + i \cosh(u) \sin(v)$.  As $\cosh(u)$ is strictly positive over the reals, this expression can only be real for $v = k \pi$ (any integral $k$, not just the evens).
This in turn requires that $\sinh(u) = \pm 1$ for even and odd $k$ respectively.
This gives you a solution set for $w$ of offset points on two vertical lines.  These extend infinitely.
$$
w = 
\begin{cases}
\operatorname{arsinh}(1) \pm k \pi i, &\text{for even } k\\
-\operatorname{arsinh}(1) \pm k \pi i, &\text{for odd } k
\end{cases}
$$
From these you can just take the inverse to get the set of points $z$ that solve the original equation.  Any of the $w$ points outside the unit circle will correspond to $z$ points inside the unit circle.
$$
z = 
\begin{cases}
(\operatorname{arsinh}(1) \pm k \pi i)^{-1}, &\text{for even } k\\
(-\operatorname{arsinh}(1) \pm k \pi i)^{-1}, &\text{for odd } k
\end{cases}
$$
As $\operatorname{arsinh(1)} \approx 0.88137$, all but the trivial solution are within the unit circle.
A: Note
$$\sinh\frac1z=i \sin\frac{-i}z=i \sin(-i\frac1z+2\pi k)=\sinh(\frac1z+i2\pi k)=1$$
Thus, $\frac1z+i2\pi k = \sinh^{-1}(1)$ and the solutions
$$z = \frac1{\sinh^{-1}(1)-i2\pi k}$$
