Define $f(z) := \frac{\cos z }{\sin z } -\frac 1z$. This is an exam-question. I have to determine singularities in $\mathbb C \cup \{ \infty \}$ and what sort of singularities they are. I further have to compute the Taylor-series around $z = 0$.
For $\cot(z)$ I see that singularities occur in $z_k = \pi k, k \in \mathbb Z$. $\cos(\pi k) \neq 0$ s.t. the $z_k$ should be poles of order one of $\cot(z)$. How must I take $\frac 1 z$ into account ? And how can I calculate the taylor series ? Is $z = 0$ a removable singularity of $f$ ?
Further question: If $\lim_{z \rightarrow 0} f(z)$ exists, is $z = 0$ then automatically a removable singularity ?
Best Answer
Since it is a simple pole, then the residue of $\cot(z)$ at $z=0$ can be determined as $$\lim_{z\to 0}z\cot(z).$$ (Hint: As $z\to 0,$ $\frac{\sin z}z\to\, ??$)
The value thus determined will be the coefficient of $z^{-1}$ in the Laurent series of $\cot(z)$ in the annulus $0<|z|<\pi$. Can you take it from here?
In answer to your other question: yes. If it were a pole, then $|f(z)|\to\infty$ as $z\to 0,$ and if essential, then $\lim_{z\to 0}|f(z)|$ fails to exist in any sense.