Term Test 2 sample P5

[Victor Ivrii]

Consider $$f(z)= \frac{16}{(z^2-16)(z^2+25)}$$ and decompose it into Laurent's series converging

(a) As $|z|<4$;

(b) As $4<|z|<5$;

(c) As $|z|>5$.

[Shengying Yang]

$$f(z)=\frac{16}{(z^2-16)(z^2+25)}$$
By doing partial fraction, we can get
$$f(z)=\frac{2}{41}\frac{1}{z-1}-\frac{2}{41}\frac{1}{z+4}-\frac{16}{41}\frac{1}{10i}(\frac{1}{z-5i}-\frac{1}{z+5i})$$
a) as $|z|<4,$
$$f(z)=\frac{2}{41}\frac{(-1)}{4}\frac{1}{1-\frac{z}{4}}-\frac{2}{41}\frac{1}{4}\frac{1}{1-\frac{(-z)}{4}}+\frac{16}{41}\frac{1}{10i}\frac{1}{5i}(\frac{1}{1-\frac{z}{5i}}+\frac{1}{1-\frac{-z}{5i}})$$
$$∴f(z)=\frac{-2}{164}(\sum_{n=0}^{\infty}(\frac{z}{4})^n+\sum_{n=0}^{\infty}(\frac{-z}{4})^n)-\frac{16}{2050}(\sum_{n=0}^{\infty}(\frac{z}{5i})^n+\sum_{n=0}^{\infty}(\frac{-z}{5i})^n)$$
Similarily,
b)as $<4|z|<5$,
$$f(z)=\frac{2}{41z}(\sum_{n=0}^{\infty}(\frac{4}{z})^n-\sum_{n=0}^{\infty}(\frac{-4}{z})^n)-\frac{16}{2050}(\sum_{n=0}^{\infty}(\frac{z}{5i})^n+\sum_{n=0}^{\infty}(\frac{-z}{5i})^n)$$
c)as $|z|>5$,
$$f(z)=\frac{2}{41z}(\sum_{n=0}^{\infty}(\frac{4}{z})^n-\sum_{n=0}^{\infty}(\frac{-4}{z})^n)-\frac{16}{410iz}(\sum_{n=0}^{\infty}(\frac{5i}{z})^n-\sum_{n=0}^{\infty}(\frac{-5i}{z})^n)$$

[Victor Ivrii]

1) You are definitely going more complicated way : one could replace $z^2$ by $w$, make a simpler decomposition and then plug $w=z^2$.

2) One need to write
$$
f(z)=\left\{\begin{aligned}
&\sum_{n=0}^\infty A_n z^n && |z|<4,\\
&\sum_{n=0}^\infty A_n z^n+\sum_{n=-\infty}^{-1} A_n z^{n},&& 4<|z|<5,\\
&\sum_{n=-\infty}^{0} A_n z^{n}&& |z|>5;
\end{aligned}\right.
$$
thus you need to collect all terms with the same power of $z$ to get a complete mark. And due to 1) it will be simpler with $z^{2n}$
(karma assigned)

[Ye Jin]

Let $w=z^2$, then $f(w)=\frac{16}{(w-16)(w+25)}$
                                 
                                  $=\frac{16}{41}\frac{1}{w-16}-\frac{16}{41}\frac{1}{w+25}$

               plug $w=z^2$, $f(z)=\frac{16}{41}\frac{1}{z^2-16}-\frac{16}{41}\frac{1}{z^2+25}$


(a) $|z|<4 ,so |w|<16 $
 
    $ f(w)=\frac{1}{41} \frac{1}{-1+\frac{w}{16}}-\frac{16}{41\cdot25} \frac{1}{1-\frac{-w}{25}}$
           
           $=\frac{-1}{41}\sum_{n=0}^{\infty}(\frac{w}{41})^n-\frac{16}{41\cdot25}\sum_{n=0}^{\infty}(\frac{-w}{25})^n$
   
          $=\sum_{n=0}^{\infty}(\frac{-1}{41\cdot16^n}-\frac{(-1)^n16}{41\cdot25^{n+1}}\,){w}^n$
   
   $f(z)=\sum_{n=0}^{\infty}(\frac{-1}{41\cdot16^n}-\frac{(-1)^n16}{41\cdot25^{n+1}}\,)z^{2n}$


(b) $4<|z|<5 ,so 16<|w|<25$
   
     $ f(w)=\frac{16}{41w} \frac{1}{1-\frac{16}{w}}-\frac{16}{41\cdot25} \frac{1}{1-\frac{-w}{25}}$

            $=\frac{16}{41w}\sum_{n=0}^{\infty}(\frac{16}{w})^n-\frac{16}{41\cdot25} \sum_{n=0}^{\infty}(\frac{-w}{25})^n$
   
            $  =\sum_{n=0}^{\infty}\frac{16^{n+1}}{41w^{n+1}}-  (\frac{(-1)^n16}{41\cdot25^{n+1}}\,){w}^n  $

            $= \sum_{n=1}^{\infty} \frac{16^n}{41w^n}-  (\frac{(-1)^n16}{41\cdot25^{n+1}}\,){w}^n  $
   
            $=\sum_{n=-\infty}^{-1} \frac{16^{-n}}{41}w^n-\sum_{n=0}^{\infty}(\frac{(-1)^n16}{41\cdot25^{n+1}}\,){w}^n  $

 $ f(z)=\sum_{n=-\infty}^{-1} \frac{16^{-n}}{41}z^{2n}-\sum_{n=0}^{\infty}(\frac{(-1)^n16}{41\cdot25^{n+1}}\,)z^{2n}  $


(c) $5<|z| ,so 25<|w|$

   $ f(w)=\frac{16}{41w} \frac{1}{1-\frac{16}{w}}-\frac{16}{41w} \frac{1}{1-\frac{-25}{w}}$

           $ =\frac{16}{41w}\sum_{n=0}^{\infty}(\frac{16}{w})^n- \sum_{n=0}^{\infty} \frac{16}{41w}(\frac{-25}{w})^n  $

           $= \sum_{n=0}^{\infty}\frac{16^{n+1}}{41w^{n+1}}-\sum_{n=0}^{\infty} \frac{16(-25)^n}{41w^{n+1}}$

           $= \sum_{n=0}^{\infty} (\frac{16^{n+1}}{41}-\frac{16(-25)^n}{41})w^{-n-1}$
   
           $= \sum_{n=-\infty}^{0} (\frac{16^{-n+1}}{41}-\frac{16(-25)^{-n}}{41})w^{n-1}$
 
    $f(z)= \sum_{n=-\infty}^{-1} (\frac{16^{-n+1}}{41}-\frac{16(-25)^{-n}}{41})z^{2n-2}$

[Victor Ivrii]

Ye, you need to check limits and powers. Obviously, as $|z|>5$ the highest power of $z$ is $-4$ and the rest are lesser

[Ye Jin]

I have modified but for |z|>5, the largest power I got is -2.

[Victor Ivrii]

Calculate the coefficient at $z^{-2}$. The answer is rather obvious because $f(z)$ decays as $z^{-4}$ as $z\to \infty$

[Ye Jin]

Calculate the coefficient at $z^{-2}$. The answer is rather obvious because $f(z)$ decays as $z^{-4}$ as $z\to \infty$

So I should start from n=-1 because the coefficient is 0 when n=0?

[Victor Ivrii]

Calculate the coefficient at $z^{-2}$. The answer is rather obvious because $f(z)$ decays as $z^{-4}$ as $z\to \infty$

So I should start from n=-1 because the coefficient is 0 when n=0?

While starting from the term with $0$ coefficient is not technically an error, it is definitely a bad bad practice.

[Ye Jin]

Calculate the coefficient at $z^{-2}$. The answer is rather obvious because $f(z)$ decays as $z^{-4}$ as $z\to \infty$

So I should start from n=-1 because the coefficient is 0 when n=0?

While starting from the term with $0$ coefficient is not technically an error, it is definitely a bad bad practice.

And now is there any other mistake I have made?