Couple things,

-Can someone please explain why f(z)=1/sin(z) is considered a simple pole and not a removable singularity?

-Also for the function,

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I know singularities exist at 2i, -2i, -3, 1. Which to me seems like there are double poles at 2i,-2i and simple poles at -3,1. However the solutions sheet indicates only -2i is a double pole and the rest are simple poles. Can anyone tell my why 2i is a simple pole??


For the first question: A pole $p$ of a function $f$ has order $n$ iff $$f(z)=\frac{\tilde{f}(z)}{(z-p)^n} $$ where $\tilde{f}$ is an holomorphic function in a disk surrounding $p$ and $n$ is the minimum integer such that this is possible. But

$$ \frac{1}{\sin{z}} =\frac{z}{\sin{z}}\cdot\frac{1}{z} \quad \text{ for }z \neq 0$$

A simple generalization of a well known formula in calculus (known sometimes, at least in Mexico, as the star limit) shows you that $z/\sin(z)$ has a well defined limit 1 when $z\to 0$. Define

$$ \tilde{f}(z)=\left\{\begin{array}{c} \frac{z}{\sin{z}} \quad z\neq 0 \\ 1, \quad z=0 \end{array}\right. $$ and compare to the definition mentioned above. Since $1/\sin{z}$ is not holomorphic at $0$, $n=1$ is the smallest for which the definition applies and so it is the order of the pole.

Alternatively: There is much simple way of proving this. It can be proved that if $f$ has $p$ as a root of multiplicity $n$ and $g$ has $p$ has a root of multiplicity $m$, then $f/g$ has $p$ as a root of multiplicity $n-m$. Where a negative multiplicity is understood as the order of a pole. So here $f=1$, $g=\sin{z}$ for which $0$ has multiplicity $1$ since $g'(0)=\cos{(0)}\neq 0$, so the multiplicity is $0-1=-1$ and the result follows.

You can apply this last technique to the other problem.

Second problem: Using the result mentioned above, see that $-2i$ is not a root of the numerator but it is a double root of the denominator, hence the result. In the $2i$ case it is a simple root of the numerator, and double root of the denominator and hence since $1-2=-1$, it is a simple root. In the other cases it is no root of the numerator (multiplicity 0) and simple root of denominator, hence order $0-1=-1$ implying simple root.

Last note: To prove the result I quoted. Use that $p$ is a root of multiplicty $m$ if $f(z)=(z-p)^m\tilde{f}(z)$ where $\tilde{f}$ is holomorphic near $p$ and $\tilde{f}(z)\neq 0$. Divide, then factor the largest power of $(z-p)$ from both numerator and denominator and then use that the quotient of holomorphic functions is holomorphic whenever the denominator does not vanishes. You obtain this useful result.

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  • $\begingroup$ I added more to the answer. $\endgroup$ – Mauricio Tec Jun 5 '14 at 2:33
  • $\begingroup$ awesome! thanks very much! $\endgroup$ – Aomine Jun 5 '14 at 2:41

For the first question, I assume you mean at $z = 0$ (or any other zero of $\sin z$). In this case, the denominator is $\sin 0 = 0$, and this is a zero of order $1$. On the other hand, the numerator has no zero, so there is no cancellation - hence the singularity is not removable.

For the second question, note that there is cancellation. The numerator has a zero of order $1$ at $z = 2i$, while the denominator has a zero of order $2$. The net effect of this is a zero of order $-1$, or a pole of order $1$.

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  • $\begingroup$ indeed i did. thank you! $\endgroup$ – Aomine Jun 5 '14 at 2:42

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