Author Topic: TT2A Problem 5  (Read 646 times)

Victor Ivrii

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TT2A Problem 5
« 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 $|z|>2$.
« Last Edit: November 29, 2018, 08:28:07 AM by Victor Ivrii »

Ye Jin

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Re: TT2A Problem 5
« Reply #1 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

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

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

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

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

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

Jingyi Wen

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Re: TT2A Problem 5
« Reply #2 on: November 24, 2018, 09:53:01 AM »
please see the attached screenshot below

Victor Ivrii

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Re: TT2A Problem 5
« Reply #3 on: November 29, 2018, 08:46:57 AM »
Ye,

please correct upper limit $n=1$ to $n=-1$ and the power of $2$ as $|z|>2$
« Last Edit: November 29, 2018, 09:21:38 AM by Victor Ivrii »

Jeffery Mcbride

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Re: TT2A Problem 5
« Reply #4 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 $\displaystyle |z|\  >\ 2$


$\displaystyle  \begin{array}{{>{\displaystyle}l}}
f( z) \ =\ \frac{2}{z} \ \frac{1}{1\ -\ \frac{2}{z}} \ +\frac{1}{z}\frac{1}{1\ +\ \frac{1}{z}} \ \\
=\ \frac{2}{z} \ \sum ^{\infty }_{n\ =\ 0}\left(\frac{2}{z}\right)^{n} +\ \frac{1}{z}\sum ^{\infty }_{n\ =\ 0}\left( -\frac{1}{z}\right)^{n}\\
\\
=\sum ^{\infty }_{n\ =\ 0}\left(\frac{2^{n+1}}{z^{n+1}}\right) \ +\ \sum ^{\infty }_{n\ =\ 0}\frac{( -1)}{z^{n+1}}^{n}\\
\\
=\sum ^{-1}_{n\ =\ -\infty }\left(\frac{1}{2^{n-2}} \ +\ \frac{1}{( -1)^{n-1}}\right) z^{n}\\
\\
\end{array}$
« Last Edit: November 29, 2018, 05:04:15 PM by Jeffery Mcbride »