Q6 TUT 0301

[Victor Ivrii]

Find the Laurent series for the given function $f(z)$ about the indicated point. Also, give the residue of the function at the point.
$$
f(z)=\frac{\sin(z)}{(z-\pi)^2};\qquad z_0=\pi.
$$

[Meng Wu]

$$\because\sin(z)=\sin(z-\pi+\pi)=-\sin(z-\pi), \space\space\space\space\space\space \space\space\space\space\text{since } \sin(\pi+\theta)=-\sin(\theta).$$
$$\therefore -\sin(z-\pi)=-\sum_{n=0}^\infty \frac{(-1)^n(z-\pi)^{2n+1}}{(2n+1)!}=\sum_{n=0}^\infty \frac{(-1)^{n+1}(z-\pi)^{2n+1}}{(2n+1)!}$$
$$\therefore \begin{align} \frac{\sin(z)}{(z-\pi)^2}= \frac{\sum_{n=0}^\infty \frac{(-1)^{n+1}(z-\pi)^{2n+1}}{(2n+1)!}}{(z-\pi)^2}=\sum_{n=0}^\infty \frac{(-1)^{n+1}(z-\pi)^{2n-1}}{(2n+1)!}\end{align}$$
$$\\$$
$$\\$$
$$\frac{\sin(z)}{(z-\pi)^2}=\frac{(-1)^{n+1}(z-\pi)^{2n-1}}{(2n+1)!}=-\frac{1}{z-\pi}+\cdots$$
Therefore, the residue of $\frac{\sin(z)}{(z-\pi)^2}$ at $z_0=\pi$ is $-1$, which is the coefficient of $(z-z_0)^{-1}$.

[Meerna Habeeb]

Set $ w=z-\pi $

So that $ z=w+\pi $

Using the identity $\sin (w+\pi )=-\sin w$

$f(w+\pi )=\frac{sin (w+\pi )}{((w+\pi )-\pi )^{2}}=\frac{-\sin
w}{w^{2}}$

$-\sin w=\sum_{n=0}^{\infty }{(-1)^{n+1} \frac{w^{2n+1}}{(2n+1)!}}$

$\frac{-\sin w}{w^{2}}=\sum_{n=0}^{\infty }{(-1)^{n+1}
\frac{w^{2n+1-2}}{(2n+1)!}}$

$\frac{-\sin w}{w^{2}}=\sum_{n=0}^{\infty }{(-1)^{n+1}
\frac{w^{2n-1}}{(2n+1)!}}$

$
-\frac{1}{w}+\frac{w}{3!}-\frac{w^{3}}{5!}+\frac{w^{5}}{7!}-\frac{w^{7}}{9!}+\ldots
$

Substitute back $w=z-\pi $

$\frac{\sin z}{(z-\pi )^{2}}=\sum_{n=0}^{\infty }{(-1)^{n+1}
\frac{(z-\pi )^{2n-1}}{(2n+1)!}}$

$=-\frac{1}{(z-\pi )}+\frac{(z-\pi )}{3!}-\frac{(z-\pi
)^{3}}{5!}+\frac{(z-\pi )^{5}}{7!}-\frac{(z-\pi )^{7}}{9!}+\ldots $

Residue of $\frac{\sin z}{(z-\pi )^{2}}$ at $z=\pi$ is -1

Which is the coefficient of $-\frac{1}{(z-\pi )}$

Or you can use $Res(f,z_{0})=\lim _{z\to z_{0}}(z-z_{0}) f(z)$

$\lim _{z\to \pi}(z-\pi ) \frac{\sin z}{(z-\pi )^{2}}=-1$

[Muyao Chen]

$\frac{sin z}{(z-π)^2}$ = $\frac{-sin (z- \pi)}{(z-π)^2}$ = $-(z - \pi)^{-2}sin (z- \pi)$
 = $-(z - \pi)^{-2} \sum_n^∞ \frac{(-1)^{n}(z- \pi)^{2n+1}}{(2n+1)!}$
= $\sum_n^∞ \frac{(-1)^{n+1}(z- \pi)^{2n-1}}{(2n+1)!}$
= $\frac{-1}{z - \pi} +\sum_k^∞ (-1)^{k+1} \frac{(z- \pi)^{2k+1}}{(2n+1)!}$
$Res(f, \pi) = -1$

[Huanglei Ln]

\begin{align*}
\frac{\sin z}{(z-\pi)^2}=\frac{-\sin(z-\pi)}{(z-\pi)^2}&=-(z-\pi)^{-2}\sin(z-\pi)\\
&=-(z-\pi)^{-2}\sum\limits_{n=0}^\infty{\frac{(-1)^n(z-\pi)^{2n+1}}{(2n+1)!}}\\
&=\sum\limits_{n=0}^\infty{\frac{(-1)^{n+1}(z-\pi)^{2n-1}}{(2n+1)!}}
\end{align*}

residue at $z_0=\pi$ is $-1$.