Example of Singularities in Complex Analysis Let $\displaystyle f(z)=\frac{z-1}{\exp(\frac{2\pi i}{z})-1}$ then,
$(1)\ \ f$ has an isolated singularity at $z=0$.
$(2)\ \ f$ has a removable singularity at $z=1$.
$(3)\ \ f$ has infinitely many poles. 
$(4)$ each pole of $f$ is of order 1.
I know all the definition of singularity.  But I can't covert this function in the simple form. Please help me thanks in advance.(here it can be more than one answers)
 A: Well, first of all, you should find all the zeroes of the denominator. This alone will allow you to draw the appropriate conclusions about $1$ and $3.$
Now, you should have found in particular that $\exp\left(\frac{2\pi i}z\right)=1$ when $z=1,$ so $$\exp\left(\frac{2\pi i}z\right)-1=(z-1)g(z)$$ for some function $g$ that's analytic in an open disk about $z=1.$ Thus, $f$ has a removable singularity at $z=1$ if and only if $g(1)\ne 0$ if and only if $\exp\left(\frac{2\pi i}z\right)-1$ has a zero of order $1$ at $z=1.$ To determine this, we differentiate both sides of $$\exp\left(\frac{2\pi i}z\right)-1=(z-1)g(z)$$ using chain and product rules to get $$-\frac{2\pi i}{z^2}\exp\left(\frac{2\pi i}z\right)=(z-1)g'(z)+g(z).$$ Letting $z=1,$ we then see that $$-2\pi i\exp(2\pi i)=g(1),$$ meaning $g(1)=-2\pi i\ne 1.$ Hence, $\exp\left(\frac{2\pi i}z\right)-1$ has a zero of order $1$ at $z=1,$ so $f(z)$ has a removable singularity at $z=1.$
Using a similar approach to the above, you can show that every zero of $\exp\left(\frac{2\pi i}z\right)-1$ is of order $1.$ What then can we say about every pole of $f(z)$?
A: Look poles at $z={1\over k},(k\in\mathbb{Z})$ has limit point $0$, so $1$ is false by definition of isolated singularity
$3,4$ true as it has infinitely many poles and each are order one clearly.
