# Thread: The order of the poles of cos(z)/z^2 and sin(z)/z^2

1. ## The order of the poles of cos(z)/z^2 and sin(z)/z^2

$\displaystyle g(z)=\frac{sin(z)}{z^2}$and $\displaystyle f(z)=\frac{cos(z)}{z^2}$ both have a pole at $\displaystyle z=0$

From the Laurent series expansion I deduced that for $\displaystyle f(z)$ this pole is of order one, while for $\displaystyle g(z)$ the pole is order two.

Was there any faster way to have noticed this (because first I thought they were of order 2 for both, since there was a square sign in the denominator....)

2. ## Re: The order of the poles of cos(z)/z^2 and sin(z)/z^2

Originally Posted by CSM
Was there any faster way to have noticed this (because first I thought they were of order 2 for both, since there was a square sign in the denominator....)

The series expansion of $\displaystyle \sin z$ and $\displaystyle \cos z$ around $\displaystyle z=0$ are well known, so we find immediately the order of the pole: $\displaystyle \sin z/z^2=1/z+\ldots$ (simple) and $\displaystyle \cos z/z^2=1/z^2+\ldots$ (double) .

3. ## Re: The order of the poles of cos(z)/z^2 and sin(z)/z^2

Originally Posted by CSM
$\displaystyle g(z)=\frac{sin(z)}{z^2}$and $\displaystyle f(z)=\frac{cos(z)}{z^2}$ both have a pole at $\displaystyle z=0$

From the Laurent series expansion I deduced that for $\displaystyle f(z)$ this pole is of order one, while for $\displaystyle g(z)$ the pole is order two.

Was there any faster way to have noticed this (because first I thought they were of order 2 for both, since there was a square sign in the denominator....)
If You know that $\displaystyle f(z)$ has a pole in $\displaystyle z=a$ but You don't know the Laurent expansion of $\displaystyle f(z)$ around $\displaystyle z=a$ , then the order of the pole in the value of n for which...

$\displaystyle \lim_{z \rightarrow a} (z-a)^{n}\ f(z)= \lambda$ (1)

... where is $\displaystyle \lambda \ne 0$. An interesting 'counterexample' is the function $\displaystyle \ln z$ , for which $\displaystyle z=0$ is a 'singularity'. If we try to apply (1) in order to determine the 'order of the pole' we discover that...

$\displaystyle \lim_{z \rightarrow 0} z^{n}\ \ln z = \begin{cases} -\infty &\phantom{|}\textrm{if }n=0\\ 0 &\phantom{|}\textrm{if } n>0 \end{cases}$ (2)

... so that no value on n satisfies (1). The reason is that $\displaystyle z=0$ is a singularity of the function $\displaystyle \ln z$ but isn't a 'pole'...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

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