Find all singularities of a function Let $f(z)=z(e^{\frac 1 z} -1)\tan{\frac 1 {z-1}}$. Find all zeros and singularities of $f$.
I know that $f$ is analytic in $\{z:z\not=0,1,\frac 1 {\pi k+\frac \pi 2}\}$ where $k\in\mathbb{Z}$ and that there isn't a isolated singularity in $z=1$ because the sequence $z_n=\frac 1 {\pi n + \frac \pi 2}\to0$ is a sequence of points that f is not analytic in any of them. I think there is an essential singularity in $z=0$, that there are simple poles in $\frac 1 {\pi n + \frac \pi 2}$ and simple zeroes in $\frac 1 {\pi n }$ when $n\not = 0$ but how can I rigorously prove that without calculating the derivative or calculating complex limits?
 A: For a meromorphic function $g$, let us write $o(f,z_0) = k$, if there is a zero-free holomorphic function $h$ in a neighbourhood $U$ of $z_0$ with
$$g(z) = (z-z_0)^k\cdot h(z)$$
on $U$. Let us call $o(f,z_0)$ the order of $f$ in $z_0$. Thus the order of $f$ is positive in zeros of $f$, negative in poles, and zero elsewhere.
Check the fundamental relations


*

*$o\left(h, z_0\right) = o(g,\varphi(z_0))$ if $h = g\circ\varphi$, and $\varphi$ is biholomorphic in a neighbourhood of $z_0$,

*$o(g\cdot h, z_0) = o(g,z_0) + o(h,z_0)$.


Then you only need to see if any two of the factors $z$, $e^{1/z}-1$, and $\tan \frac{1}{z-1}$ have any common "special points", and if so, add their orders to find out what kind of point the product has.
That deals with the multiplicities of the zeros and poles.
For the essential singularity of $e^{1/z}-1$ in $0$, note that the only way to cancel an essential singularity in a product (except a factor $\equiv 0$) is an essential singularity in another factor (a product/quotient of poles or zeros can only ever result in a pole, or a removable singularity), but the other factors don't have essential singularities in $0$, because the poles of $\tan \frac{1}{z-1}$ occur in $z_k = 1 + \frac{1}{(k+1/2)\pi}$, not in $\frac{1}{(k+1/2)\pi}$ as you wrote, similar for the zeros.
And take care of the non-isolated singularities, these may be classified as essential, or remain unsubsumed if the term "essential singularity" is reserved for isolated singularities.
A: Notice that there is an error in your post. The poles of $\tan\frac{1}{z-1}$ do not have $z=0$ as a limit point. Their limit point is actually $z=1$.
A function $\mathbb{f}$ is said to have an essential singularity at $z=0$ if $z=0$ is a singularity of $\mathbb{f}$, the restriction of $\mathrm{f}$ to a sufficiently small annulus $0 < |z| < \varepsilon$ is meromorphic and $z=0$ is neither a pole nor a removable singularity.
First of all, if $0 < |z| < \tfrac{1}{2}$, then $z$ and $\mathrm{e}^{1/z}-1$ are holomorphic. Moreover, and most importantly, all of the singularities of $\tan\frac{1}{z-1}$ are isolated: the distances between the simple poles are all finite. It follows that $\mathrm{f}$ is meromorphic on the open annulus $0 < |z| < \tfrac{1}{2}$.
Next, you should notice that $\mathrm{e}^{1/z}$ has an essential singularity at $z=0$. (To see this, notice that $\mathrm{e}^z = 1 + z + \tfrac{1}{2!}z^2 + \tfrac{1}{3!}z^3 + \cdots $, and so $\mathrm{e}^{1/z} = 1 + z^{-1} + \tfrac{1}{2!}z^{-2} + \tfrac{1}{3!}z^{-3} + \cdots$ for all $z \neq 0$). If the $\tan$-term wasn't included then your function would have an essential singularity at $z=0$.
It follows that $z=0$ is an essential singularity of your function.
