# Toronto Math Forum

## MAT334-2018F => MAT334--Lectures & Home Assignments => Topic started by: Kathy Ngo on November 20, 2018, 12:18:52 AM

Title: 2.5 - Q23
Post by: Kathy Ngo on November 20, 2018, 12:18:52 AM
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.
Title: Re: 2.5 - Q23
Post by: Amy Zhao on November 20, 2018, 11:22:03 AM
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
Title: Re: 2.5 - Q23
Post by: Min Gyu Woo on November 20, 2018, 07:02:57 PM
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.
Title: Re: 2.5 - Q23
Post by: Ende Jin on November 20, 2018, 08:24:14 PM
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.
Title: Re: 2.5 - Q23
Post by: Min Gyu Woo on November 20, 2018, 10:05:16 PM
Is it just me or is the entirety of Q23 answer key wrong...
Title: Re: 2.5 - Q23
Post by: Amy Zhao on November 20, 2018, 10:40:57 PM
I was also wondering if the answer key is wrong.
I don't understand how to get the part $a_n=\frac{-4}{3^{n+2}}$.
Title: Re: 2.5 - Q23
Post by: Victor Ivrii on November 21, 2018, 05:11:49 AM
Ende
Fix  parenthesis