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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?
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
$\chi$ $\sigma$
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.