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In summary: 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

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Dustinsfl

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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?

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chisigma

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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?

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

$\chi$ $\sigma$

- #3

Dustinsfl

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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

$\chi$ $\sigma$

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

- #4

chisigma

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dwsmith said:

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

$\chi$ $\sigma$

- #5

blue_raver22

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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.

## 1. What is the definition of "Residue"?

The residue of a function at a point is the coefficient of the term with the highest negative power in its Laurent series expansion around that point.

## 2. How is the residue calculated?

The residue can be calculated by taking the limit as z approaches the point of interest of the function multiplied by (z - point of interest)^n, where n is the highest negative power in the Laurent series expansion.

## 3. What is the significance of the residue in complex analysis?

The residue is important in the calculation of complex integrals, as it allows for the evaluation of integrals involving singularities such as poles. It also has applications in physics, engineering, and other fields.

## 4. Can the residue of \frac{\sin z}{z^n} be negative?

Yes, the residue can be negative if the highest negative power in the Laurent series expansion is an odd integer. In this case, the residue would have a negative coefficient.

## 5. Is the residue of \frac{\sin z}{z^n} always non-zero?

No, the residue can be zero if the function has a removable singularity or if the highest negative power in the Laurent series expansion is zero. In these cases, the residue would be equal to zero.

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