MAT334-2018F > Reading Week Bonus--sample problems for TT2

Term Test 2 sample P5

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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