2.5 - Q23

[Kathy Ngo]

Can anyone show me how they did part (a)?
I got a series but it's not the same as the solutions in the back of the textbook.

[Amy Zhao]

Here is what I did for part (a):
For $1<|z|<2$, if we follow equation (12) and (13), we have
$Res(\frac{z+2}{\frac{z-2}{z+1}};-1) = \frac{z+2}{z-2}|_{z=-1}= -\frac{1}{3}$

$Res(\frac{z+2}{\frac{z+1}{z-2}};2) = \frac{z+2}{z+1}|_{z=2}= \frac{4}{3}$

By (12) and (13):
$\sum_{-\infty}^{\infty}a_nz^n$,where
$a_n = \sum\frac{1}{3}(-1)^n,n=-1,-2,...$
$and$
$a_n = -\sum\frac{4}{3}(2)^{-n-1}=-\sum\frac{4}{3}(\frac{1}{2^{n+1}}),n=0,1,2,...$

Similar for when $2<|z|<\infty$
You can read example 12 from 2.5

[Min Gyu Woo]

Can you guys explain why you can use (12) for z = -1, and (13) for z=2?

Following the definition in the textbook you can't use either (12) or (13) because |z_j| < r and |z_j| > R aren't satisfied.

[Ende Jin]

RE: Min Gyu Woo

If you only want a legitimate procedure for the answer to the question, you can just "instantiate" the proof.
Since
\begin{align*}
    \frac{z + 2}{ (z-2) (z + 1)} & = \frac{4}{3}(\frac{1}{2} \frac{-1}{1 - \frac{z}{2}}) - \frac{1}{3} (\frac{1}{z}\frac{1}{1+\frac{1}{z}}) \\
    &= - \frac{4}{3} \frac{1}{2} (\sum (\frac{z}{2})^n) - \frac{1}{3} \frac{1}{z}(\sum (\frac{1}{z})^n)
\end{align*}

The second line is correct because the convergence relies on the geometric series instead of (12) and (13).

However, it is reasonable to ask if (12) and (13) can be extended to less than or equal to.

The second part of the question can solve similarly.

[Min Gyu Woo]

Is it just me or is the entirety of Q23 answer key wrong...

[Amy Zhao]

I was also wondering if the answer key is wrong.
I don't understand how to get the part $a_n=\frac{-4}{3^{n+2}}$.

[Victor Ivrii]

Ende
Fix  parenthesis