Author Topic: TT2B Problem 5  (Read 298 times)

Victor Ivrii

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TT2B Problem 5
« on: November 24, 2018, 05:22:35 AM »
Consider $$f(z)= \frac{8}{(z-3)(z+5)}$$ and decompose it into Laurent's series converging

(a) As $|z|<3$;

(b) As $3<|z|<5$;

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

Shengying Yang

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Re: TT2B Problem 5
« Reply #1 on: November 24, 2018, 08:34:22 AM »
$$f(z)=\frac{8}{(z-3)(z+5)}=\frac{1}{z-3}-\frac{1}{z+5}$$
(a)as$|z|<3$,
$$f(z)= -\frac{1}{3} \frac{1}{1-\frac{z}{3}}- \frac{1}{5} \frac{1}{1-(-\frac{z}{5})}=-\frac{1}{3}\sum_{n=0}^{\infty}(\frac{z}{3})^n-\frac{1}{5}\sum_{n=0}^{\infty}(-\frac{z}{5})^n$$
(b)as$3<|z|<5$,
$$f(z)= \frac{1}{z} \frac{1}{1-\frac{3}{z}}- \frac{1}{5} \frac{1}{1-(-\frac{z}{5})}=\frac{1}{z}\sum_{n=0}^{\infty}(\frac{3}{z})^n-\frac{1}{5}\sum_{n=0}^{\infty}(-\frac{z}{5})^n=\sum_{n=0}^{\infty}3^nz^{-n-1}-\frac{1}{5}\sum_{n=0}^{\infty}(-\frac{z}{5})^n$$
(c)as$|z|>5$,
$$f(z)= \frac{1}{z} \frac{1}{1-\frac{3}{z}}- \frac{1}{z} \frac{1}{1-(-\frac{5}{z})}=\sum_{n=0}^{\infty}3^nz^{-n-1}-\sum_{n=0}^{\infty}(-5)^nz^{-n-1}=\sum_{n=0}^{\infty}z^{-n-1}(3^n-(-5)^n)$$

« Last Edit: November 24, 2018, 08:59:03 AM by Shengying Yang »

Muyao Chen

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Re: TT2B Problem 5
« Reply #2 on: November 24, 2018, 11:23:38 AM »
$$f(z) = \frac{1}{z-3} - \frac{1}{z-5}$$

(a) $\mid z \mid$ $<$ 3

$$ f(z) = \frac{1}{3} - \frac{1}{1 - \frac{z}{3}} - \frac{1}{5} \frac{1}{1+ \frac{z}{5}}   
 = - \frac{1}{3} \sum_{n=0}^{\infty} \frac {z}{3}^{n} -
\frac{1}{5}  \sum_{n=0}^{\infty} - \frac {z}{5}^{n}
 =  \sum_{n=0}^{\infty} (- \frac{1}{3^{n +1}} - \frac{(-1)^{n}}{5^{n+1}}) z^{n} $$

(b) 3 $<$ $\mid z \mid$ $<$ 5

$$f(z) = \frac{1}{z}  \frac{1}{1 - \frac{3}{z}} - \frac{1}{5} \frac{1}{1+ \frac{z}{5}}

 =   \frac{1}{z} \sum_{n=0}^{\infty} \frac {3}{z}^{n} -
\frac{1}{5}  \sum_{n=0}^{\infty} - \frac {z}{5}^{n} 
=  \sum_{n=0}^{\infty}  \frac{3^{n}}{z^{n+1}} - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{5^{n+1}}z^{n}
=  \sum_{n = - \infty}^{1} 3^{-n} z^{n-1} - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{5^{n+1}}z^{n}$$

(c) $\mid z \mid$ $>$ 5

$$f(z) = \frac{1}{z}  \frac{1}{1 - \frac{3}{z}} - \frac{1}{z} \frac{1}{1+ \frac{5}{z}}
= \frac{1}{z} \sum_{n=0}^{\infty} \frac {3}{z}^{n} -
\frac{1}{z}  \sum_{n=0}^{\infty} - \frac {5^{n}}{z^{n}} 
= \sum_{n=0}^{\infty}  (\frac{3^{n}}{z^{n+1}} - (-1) ^{n} \frac{5^{n}}{z^{n+1}})
= \sum_{n = - \infty}^{0}  (3^{-n} - (-1)^{-n} 5^{-n} ) z^{n-1}$$

« Last Edit: November 24, 2018, 11:30:55 AM by Muyao Chen »

Victor Ivrii

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Re: TT2B Problem 5
« Reply #3 on: December 03, 2018, 07:48:36 PM »
Some misprints