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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 ?

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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.

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  • $\begingroup$ Ok nice. I didn't knew this. So the residue of $\cot(z)$ will be $1$ s.t. $\frac 1 z$ cancel out in $f$. Must I then define $f(0) := \lim_{z \rightarrow 0}f(z) $ and compute the Taylor Series via the $n-$th derivatives ? $\endgroup$
    – user42761
    Jun 28, 2013 at 14:26
  • $\begingroup$ The Taylor series of the "repaired" function will be the Laurent series of $f(z)$--as originally defined--with the singular part removed. This Taylor series will provide the definition--no need to state that $f(0):=\lim_{z\to 0}f(z)$. $n$th derivatives of $\cot(z)$ are probably the way to go, but there may be a more tractable method that isn't occurring to me. $\endgroup$ Jun 28, 2013 at 14:32

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