Help identifying the singularities of $\csc(\cos z) = \frac{1}{\sin(\cos z)}$ I am really stuck with this one:
$\frac{1}{\sin(\cos z)}$ has a singularity when $\cos z = k\pi $ since $\sin(k\pi) = 0$ but how do I solve for the value of $z$, how can i evaluate $\cos z = k\pi $?
 A: Solve $\cos z = w$, let $y=e^{iz}$. Then $\cos z = \frac 12\left(y+y^{-1}\right)$. So $y+\frac{1}{y} = 2w$, or $y^2-2wy+1=0$ or $$y=\frac{2w\pm\sqrt{4w^2-4}}{2}= w \pm \sqrt{w^2-1}$$
So $iz = \log\left(k\pi\pm \sqrt{k^2\pi^2-1}\right)$.
So:
$$z = i\log\left(k\pi\pm \sqrt{k^2\pi^2-1}\right)$$
Since $(k\pi+ \sqrt{k^2\pi^2-1})(k\pi- \sqrt{k^2\pi^2-1})$=1, you actually have:
$$z=\pm i\log\left(k\pi+ \sqrt{k^2\pi^2-1}\right)$$
Now, $\log$ is a multivalued function, so for each $k$ you get infinitely many poles.
When $k=0$, the roots are $z=\frac{\pi}{2}+m\pi$ for $m\in\mathbb Z$. Those are the "real" roots.
When $k>0$, $k\pi+\sqrt{k^2\pi^2-1}$ is positive real, so $z$ is necessarily an imaginary number plus some multiply of $2\pi$.
When $k<0$, we can take any pole for $-k>0$ and add $\pi$. 
So for $k>0$, the polls corresponding to $k$ and $-k$ are:
$$\pm i\ln\left(k\pi+ \sqrt{k^2\pi^2-1}\right) + m\pi$$ for some $m\in\mathbb Z$, and $\ln$ is the standard real natural log.
A: Rough answer.
$$\cos z=\frac{e^{iz}+e^{-iz}}{2}$$
So we must solve:
$$\frac{e^{iz}+e^{-iz}}{2}=\pi k.$$
After some algebra, we get
$$e^{2iz}-2\pi ke^{iz}+1=0.$$
Let $w=e^{iz}$, the equation is now quadratic in $w$. So, by quadratic formula,
$$w=\pi k\pm\sqrt{\pi^2k^2-1}.$$
Substituting $e^{iz}$ back for w and solving for $z$ gives
$$z=-i\log\left(\pi k\pm\sqrt{\pi^2k^2-1}\right).$$
Assuming that we are using the principal branch of $\log$, you will have to add all integer multiples of $2\pi i$ to the final answer.
