Singularity of $\log\left(1 - \frac{1}{z}\right)$ I am looking for the classification of the all singularity of the function $\log\left(1 - \frac{1}{z}\right)$.
I know $\log\left(1 - \frac{1}{z}\right) = \log(z -1) -\log(z)$. This implies that $z =0 , 1$ are the points of singularity. But I am not able to classify them i.e. $z= 0,1$ are pole or removable singularity or essential singularity.
Please help me with this problem .
Thanks.
 A: One must be careful here. We can define $\log(z)$ for $z\in\mathbb{C}$ if we leave out the negative real axis. It is this branch cut that causes problems.
Riemann's Theorem on removable singularities says that if $f$ is defined on a punctured neighborhood of $a$ and
$$
\lim_{z\to a}(z-a)f(z)=0
$$
then $f$ has a removable singularity at $a$. Indeed, if we consider
$$
\left|z\log(z)\right|\le\left|z\right|(\left|\log|z|\right|+\pi)
$$
and
$$
\lim_{t\to0^+}t\log(t)=0
$$
we get that
$$
\lim_{z\to0}z\log(z)=0
$$
for all $z\in\mathbb{C}$ as long as $\boldsymbol{z}$ is not on the negative real axis. The fact that $z\log(z)$ is not defined on a full punctured neighborhood of $0$, means that although we have the limit above, Riemann's Theorem does not apply.
This is the problem with $\log\left(1-\frac1z\right)$; we cannot define this function in a punctured neighborhood of $0$ or $1$. These are branch points of this function.
A: First you don't know that $$\log\left(1 - \frac{1}{z}\right) = \log(z -1) -\log(z).$$ This property is true for real numbers not in general for complex numbers. You must chose a determination for your logarithm, I guess you cut $]-\infty,0].$ Hence you must find the complex numbers $z$ such that $$1-\frac{1}{z} \in ]-\infty,0].$$ It is easily seen that this equivalent to $$z \in ]0,1].$$ These are no isolated singularities. Moreover you can't divide by zero, hence $0$ is an other singularity. To know what kind of singularity it is, simply look at the limit $$\lim_{z \to 0} \log\left(1 - \frac{1}{z}\right) = \infty.$$ This implies that zero is a pole. (Removable singularities have finite limits and essential singularities don't have limits). Finally to determine the order of the pole, you can for example use the Laurent's expansion. 
