How to Calculate the Residue of \(\frac{\sin z}{z^n}\)?

[Dustinsfl]

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]

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?

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

$\chi$ $\sigma$

 

[Dustinsfl]

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

$\chi$ $\sigma$

Is there another way to do this? I haven't seen that definition of residue before.

 

[chisigma]

Is there another way to do this? I haven't seen that definition of residue before.

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...

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

$\chi$ $\sigma$

 

[blue_raver22]

I would suggest approaching this problem by using the Cauchy Residue Theorem. This theorem states that the residue of a function at a certain point is equal to the coefficient of the \frac{1}{z} term in the Laurent series expansion of the function at that point. In this case, the function is \frac{\sin z}{z^n} and we are trying to find the residue at a certain point.

To start, we can use the Weierstrass product definition of \sin z to write the function as a product of infinite factors. Then, we can expand each factor using the Taylor series expansion of \sin z. From there, we can combine the terms and group them according to powers of z. This will give us the Laurent series expansion of the function at the given point.

Once we have the Laurent series, we can easily identify the coefficient of the \frac{1}{z} term, which will be the residue of the function at that point. This process can be repeated for different values of n to find all the residues of \frac{\sin z}{z^n}.

In summary, using the Cauchy Residue Theorem and the Weierstrass product definition of \sin z, we can find the residues of \frac{\sin z}{z^n} at any given point. This method is efficient and reliable in finding the residues of complex functions.