Calculating Laurent series of complex functions

[MatinSAR]

Homework Statement

Calculate laurent series of following function.

Relevant Equations

series expansion formulas.

1. $f(z)=\dfrac {\sin z}{z- \pi}$ at $z=\pi$ : $$ \dfrac {\sin z}{z- \pi}=\dfrac {\sin(\pi +z- \pi)}{z- \pi}=\dfrac {- \sin(z- \pi)}{z- \pi}=\dfrac {-1}{z- \pi} \sum_{n=0}^\infty \dfrac {(-1)^n (z- \pi)^{2n}}{(2n+1)!}$$My answer has extra $\dfrac {-1}{z- \pi}$ according to a calculator. Am I wrong?

2. Find the residue of $f(z)=\dfrac {\sin z} {z^4}$ at $z=0$ :
I wrote laurent series ot this. It doesn't have $1/z$. I think the residue is $0$. Is it it true?

3. Calculate the residue of the $f(z)=z^2 \sin \dfrac {1}{z+1}$ at its singular points.
I'm not sure what should I do. Should I expand both $z^2$ and $\sin \dfrac {1}{z+1}$ at $z=-1$?

Any help would be appreciated.

 

[Orodruin]

1. The sine expansion should have $(z-\pi)^{2n+1}$, not $(z-\pi)^{2n}$

2. No. The sine has all odd $z$ in the expansion so in particular the $z^3$ term from the sine becomes a 1/z term when divided by $z^4$.

 

[MatinSAR]

1. The sine expansion should have $(z-\pi)^{2n+1}$, not $(z-\pi)^{2n}$

It's clear now.

2. No. The sine has all odd $z$ in the expansion so in particular the $z^3$ term from the sine becomes a 1/z term when divided by $z^4$.

There should be a mistake in my expansion. I will try again.


Thank you for your time.

 

[MatinSAR]

1. The sine expansion should have $(z-\pi)^{2n+1}$, not $(z-\pi)^{2n}$

2. No. The sine has all odd $z$ in the expansion so in particular the $z^3$ term from the sine becomes a 1/z term when divided by $z^4$.

$$ f(z) = \dfrac{\sin(z)}{z^4} = \dfrac{z - \dfrac{z^3}{3!} + \dfrac{z^5}{5!} - \dfrac{z^7}{7!} + \cdots}{z^4} $$Therefore, the residue of $f(z) = \frac{\sin(z)}{z^4}$ at $z = 0$ is $-\frac{1}{6}$. Thank you for your help.