Toronto Math Forum

MAT334-2018F => MAT334--Tests => Term Test 2 => Topic started by: Victor Ivrii on November 24, 2018, 05:17:20 AM

Title: TT2A Problem 5
Post by: Victor Ivrii on November 24, 2018, 05:17:20 AM
Consider $$f(z)= \frac{3z}{(z-2)(z+1)}$$ and decompose it into Laurent's series converging

(a) As $|z|<1$;

(b) As $1<|z|<2$;

(c) As $|z|>2$.
Title: Re: TT2A Problem 5
Post by: Ye Jin on November 24, 2018, 08:35:43 AM
$f(z)=\frac{2}{z-2}+\frac{1}{z+1}$

(a) |z|<1

$f(z)= \frac{1}{\frac{z}{2}-1}$+$\frac{1}{z+1}$

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

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

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

(b) 1<|z|<2

$f(z)= \frac{1}{\frac{z}{2}-1}$+$\frac{1}{z}\frac{1}{1+\frac{1}{z}}$

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

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

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

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

(c)|z|>2

$f(z)= \frac{2}{z}\frac{1}{1-\frac{2}{z}}$+$\frac{1}{z}\frac{1}{1+\frac{1}{z}}$

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

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

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

$=\sum_{n=-\infty}^{1}(2^{n+2}+(-1)^{n+1})z^{n}$
Title: Re: TT2A Problem 5
Post by: Jingyi Wen on November 24, 2018, 09:53:01 AM
please see the attached screenshot below
Title: Re: TT2A Problem 5
Post by: Victor Ivrii on November 29, 2018, 08:46:57 AM
Ye,

please correct upper limit $n=1$ to $n=-1$ and the power of $2$ as $|z|>2$
Title: Re: TT2A Problem 5
Post by: Jeffery Mcbride on November 29, 2018, 05:00:07 PM
$\displaystyle f( z) \ =\ \frac{3z}{( z-2)( z+1)}$

By partial fractions:

$\displaystyle \begin{array}{{>{\displaystyle}l}} f( z) \ =\ \frac{A}{z-2} \ +\ \frac{B}{z+1}\\ \\ f( z) \ =\ \frac{2}{z-2} \ +\ \frac{1}{z+1} \end{array}$

a) as $\displaystyle |z|\ < \ 1$

$\displaystyle \begin{array}{{>{\displaystyle}l}} f( z) \ =\ -\frac{1}{1\ -\ \frac{z}{2}} \ +\ \frac{1}{1\ -\ ( -z)}\\ \\ =\ -\sum ^{\infty }_{n\ =\ 0}\left(\frac{z}{2}\right)^{n} \ +\ \sum ^{\infty }_{n\ =\ 0}( -z)^{n}\\ \\ =\sum ^{\infty }_{n\ =\ 0}\left( -\frac{1}{2^{n}} \ +( -1)^{n}\right) z^{n} \end{array}$

b) as $\displaystyle 1\ < \ |z|\ < \ 2$

$\displaystyle f( z) \ =\$$\displaystyle -\frac{1}{1\ -\ \frac{z}{2}} \ +\ \frac{1}{z}\frac{1}{1\ +\ \frac{1}{z}}$

$\displaystyle \begin{array}{{>{\displaystyle}l}} =-\sum ^{\infty }_{n\ =\ 0}\left(\frac{z}{2}\right)^{n} \ +\ \frac{1}{z}\sum ^{\infty }_{n\ =\ 0}\left( -\frac{1}{z}\right)^{n}\\ \\ =-\sum ^{\infty }_{n\ =\ 0}\frac{z^{n}}{2^{n}} \ +\ \sum ^{\infty }_{n\ =\ 0}\frac{( -1)}{z^{n+1}}^{n}\\ \\ =\sum ^{\infty }_{n\ =\ 0} -\frac{1}{2^{n}} \ z^{n} \ +\ \sum ^{1}_{n\ =\ -\infty }\frac{1}{( -1)^{n-1}} \ z^{n} \end{array}$

c) as $\displaystyle |z|\ >\ 2$

$\displaystyle \begin{array}{{>{\displaystyle}l}} f( z) \ =\ \frac{2}{z} \ \frac{1}{1\ -\ \frac{2}{z}} \ +\frac{1}{z}\frac{1}{1\ +\ \frac{1}{z}} \ \\ =\ \frac{2}{z} \ \sum ^{\infty }_{n\ =\ 0}\left(\frac{2}{z}\right)^{n} +\ \frac{1}{z}\sum ^{\infty }_{n\ =\ 0}\left( -\frac{1}{z}\right)^{n}\\ \\ =\sum ^{\infty }_{n\ =\ 0}\left(\frac{2^{n+1}}{z^{n+1}}\right) \ +\ \sum ^{\infty }_{n\ =\ 0}\frac{( -1)}{z^{n+1}}^{n}\\ \\ =\sum ^{-1}_{n\ =\ -\infty }\left(\frac{1}{2^{n-2}} \ +\ \frac{1}{( -1)^{n-1}}\right) z^{n}\\ \\ \end{array}$