The discussion focuses on calculating the residue of the function \(\frac{\sin z}{z^n}\) at the pole \(z=0\). The residue can be computed using the formula \(r= \frac{1}{(n-1)!}\ \lim_{z \rightarrow 0} \frac{d^{n-1}}{d z^{n-1}} \sin z\). An alternative method involves using the Maclaurin series expansion of \(\sin z\) and identifying the coefficient of the \(\frac{1}{z}\) term in the Laurent series of \(\frac{\sin z}{z^n}\). For even \(n\), the residue is given by \(r_{n}= \frac{(-1)^{\frac{n}{2}-1}}{(n-1)!}\), while for odd \(n\), the residue is zero.
PREREQUISITESMathematicians, students of complex analysis, and anyone interested in advanced calculus techniques for evaluating residues in complex functions.
dwsmith said: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?
chisigma said: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
$\chi$ $\sigma$
dwsmith said:Is there another way to do this? I haven't seen that definition of residue before.