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- Thread starter dwsmith
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- Feb 13, 2012

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The function has a pole of order n in z=0, so that its residue is...I am tying to find all the residue of $\dfrac{\sin z}{z^n}$.

I am think can I do this but I am not sure where to start. Should I use the Weierstrass product definition of sin z?

$\displaystyle r= \frac{1}{(n-1)!}\ \lim_{z \rightarrow 0} \frac{d^{n-1}}{d z^{n-1}} z^{n}\ f(z)$ (1)

... and in Your case is...

$\displaystyle r= \frac{1}{(n-1)!}\ \lim_{z \rightarrow 0} \frac{d^{n-1}}{d z^{n-1}} \sin z$ (2)

... which is very easy to compute...

Kind regards

$\chi$ $\sigma$

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Is there another way to do this? I haven't seen that definition of residue before.The function has a pole of order n in z=0, so that its residue is...

$\displaystyle r= \frac{1}{(n-1)!}\ \lim_{z \rightarrow 0} \frac{d^{n-1}}{d z^{n-1}} z^{n}\ f(z)$ (1)

... and in Your case is...

$\displaystyle r= \frac{1}{(n-1)!}\ \lim_{z \rightarrow 0} \frac{d^{n-1}}{d z^{n-1}} \sin z$ (2)

... which is very easy to compute...

Kind regards

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

What You say is a little surprising... but never mind, there is a different way to arrive to the same result. If You write the McLaurin expansion of the sine function...Is there another way to do this? I haven't seen that definition of residue before.

$\displaystyle \sin z =\sum_{k=1}^{\infty} \frac{(-1)^{k}}{(2k+1)!}\ z^{2k+1}$ (1)

... then You divide $\sin z$ by $z^{n}$ and search the coefficient of the term in $\frac{1}{z}$ in the Laurent expansion of $\displaystyle \frac{\sin z}{z^{n}}$ You obtain for n even and $n>)$...

$\displaystyle r_{n}= \frac{(-1)^{\frac{n}{2}-1}}{(n-1)!}$ (2)

... and for any other value of n $r_{n}=0$...

Kind regards

$\chi$ $\sigma$