find laurant series lf (1)f(z)=(1/(z−1))+(1/(z+2)) ,∣z∣>1 (2)f(z)=(1/(z−2))−(2/(z−3)) ,∣z∣<3


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Question Number 147066 by tabata last updated on 17/Jul/21

$f i n d l a u r a n t s e r i e s l f$ $(1) f (z) = \frac{1}{z - 1} + \frac{1}{z + 2}, ∣ z ∣> 1$ $(2) f (z) = \frac{1}{z - 2} - \frac{2}{z - 3}, ∣ z ∣< 3$
	Answered by mathmax by abdo last updated on 17/Jul/21

	$1) f (z) = \frac{1}{z - 1} + \frac{1}{z + 2} we have \frac{1}{z - 1} = \frac{1}{z (1 - \frac{1}{z})} (∣ \frac{1}{z} ∣< 1)$ $= \frac{1}{z} \sum_{n = 0}^{\infty} \frac{1}{z^{n}} = \sum_{n = 0}^{\infty} \frac{1}{z^{n + 1}} = \sum_{n = 1}^{\infty} \frac{1}{z^{n}}$ $\frac{1}{z + 2} =_{\frac{1}{z} = y} \frac{1}{\frac{1}{y} + 2} = \frac{y}{1 + 2 y}$ $2 y = \frac{2}{z} \Rightarrow∣ 2 y ∣= \frac{2}{∣ z ∣} < 1 \Rightarrow \frac{∣ z ∣}{2} > 1 \Rightarrow∣ z ∣> 2 we get$ $\frac{y}{1 + 2 y} = y \sum_{n = 0}^{\infty} {(- 1)}^{n} {(2 y)}^{n} = \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{2^{n}}{z^{n}} \Rightarrow$ $f (z) = \sum_{n = 1}^{\infty} \frac{1}{z^{n}} + \sum_{n = 0}^{\infty} \frac{{(- 2)}^{n}}{z^{n}} = 1 + \sum_{n = 1}^{\infty} (1 + {(- 2)}^{n}) \frac{1}{z^{n}}$ $∣ 2 y ∣> 1 \Rightarrow \frac{2}{∣ z ∣} > 1 \Rightarrow 1 <∣ z ∣< 2 \Rightarrow$ $\frac{y}{1 + 2 y} = \frac{y}{2 y (1 + \frac{1}{2 y})} = \frac{1}{2} \sum_{n = 0}^{\infty} {(- 1)}^{n} {(\frac{1}{2 y})}^{n}$ $= \frac{1}{2} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2^{n} y^{n}} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2^{n + 1}} z^{n} \Rightarrow$ $f (z) = \sum_{n = 1}^{\infty} \frac{1}{z^{n}} + \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2^{n + 1}} z^{n}$
	Answered by Olaf_Thorendsen last updated on 17/Jul/21

	$(1)$ $f (z) = \frac{1}{z - 1} + \frac{1}{z + 2}, ∣ z ∣> 1$ $f (z) = \frac{1}{z} . \frac{1}{1 - \frac{1}{z}} + \frac{1}{2} . \frac{1}{1 + \frac{z}{2}}$ $∣ z ∣> 1 \Rightarrow ∣ \frac{1}{z} ∣< 1$ $f (z) = \frac{1}{z} . \underset{n = 0}{\sum^{\infty}} \frac{1}{z^{n}} + \frac{1}{2} . \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{z^{n}}{2^{n}}$ $f (z) = \underset{n = 1}{\sum^{\infty}} \frac{1}{z^{n}} + \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{z^{n}}{2^{n + 1}}$ $(2)$ $f (z) = \frac{1}{z - 2} - \frac{2}{z - 3}, ∣ z ∣< 3$ $- \frac{2}{z - 3} = \frac{2}{3} . \frac{1}{1 - \frac{z}{3}} = - \frac{2}{3} \underset{n = 0}{\sum^{\infty}} \frac{z^{n}}{3^{n}}$ $1 st case : ∣ z ∣< 2$ $\frac{1}{z - 2} = - \frac{1}{2} . \frac{1}{1 - \frac{z}{2}} = - \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \frac{z^{n}}{2^{n}}$ $f (z) = - \underset{n = 0}{\sum^{\infty}} (\frac{1}{2^{n + 1}} + \frac{2}{3^{n + 1}}) z^{n}$ $2 nd case : 2 <∣ z ∣< 3 \Rightarrow \frac{2}{∣ z ∣} < 1$ $\frac{1}{z - 2} = \frac{1}{z} . \frac{1}{1 - \frac{2}{z}} = \frac{1}{z} \underset{n = 0}{\sum^{\infty}} \frac{2^{n}}{z^{n}} = \underset{n = 1}{\sum^{\infty}} \frac{2^{n - 1}}{z^{n}}$ $f (z) = \underset{n = 1}{\sum^{\infty}} \frac{2^{n - 1}}{z^{n}} - \underset{n = 0}{\sum^{\infty}} \frac{2}{3^{n + 1}} z^{n}$
	Commented byMrsof last updated on 17/Jul/21

	$s i r w h y y o u g i v e ∣ z ∣< 2 ?$
	Commented byOlaf_Thorendsen last updated on 17/Jul/21

	$b e c a u s e i n t h e f i r s t c a s e t h e s e r i e$ $i s c o n v e r g e n t o n l y f o r ∣ z ∣< 2$
	Commented byMrsof last updated on 18/Jul/21

	$t h a n k y o u s i r$
	Answered by mathmax by abdo last updated on 17/Jul/21

	$2) g (z) = \frac{1}{z - 2} - \frac{2}{z - 3} we have$ $\frac{2}{z - 3} = - \frac{2}{3 - z} = \frac{- 2}{3 (1 - \frac{z}{3})} = - \frac{2}{3} \sum_{n = 0}^{\infty} \frac{z^{n}}{3^{n}} = - 2 \sum_{n = 0}^{\infty} \frac{z^{n}}{3^{n + 1}} (∣ \frac{z}{3} ∣< 1)$ $\frac{1}{z - 2} =_{y = \frac{z}{3}} \frac{1}{3 y - 2} = \frac{- 1}{2 - 3 y} = \frac{- 1}{2 (1 - \frac{3}{2} y)}$ $∣ \frac{3 y}{2} ∣=∣ \frac{z}{2} ∣< 1 \Rightarrow∣ z ∣< 2 inthis case we get$ $\frac{1}{z - 2} = - \frac{1}{2} \sum_{n = 0}^{\infty} {(\frac{3}{2})}^{n} y^{n} = - \frac{1}{2} \sum_{n = 0}^{\infty} \frac{3^{n}}{2^{n}} \frac{z^{n}}{3^{n}} = - \frac{1}{2} \sum_{n = 0}^{\infty} \frac{z^{n}}{2^{n}} \Rightarrow$ $g (z) = - 2 \sum_{n = 0}^{\infty} \frac{z^{n}}{3^{n + 1}} - \sum_{n = 0}^{\infty} \frac{z^{n}}{2^{n + 1}} = \sum_{n = 0}^{\infty} (- \frac{2}{3^{n + 1}} - \frac{1}{2^{n + 1}}) z^{n}$ $∣ \frac{3 y}{2} ∣> 1 \Rightarrow∣ \frac{z}{2} ∣> 1 \Rightarrow 2 <∣ z ∣< 3 we get$ $\frac{1}{z - 2} = \frac{- 1}{3 y (\frac{2}{3 y} - 1)} = \frac{1}{3 y (1 - \frac{2}{3 y})} = \frac{1}{3 y} \sum_{n = 0}^{\infty} {(\frac{2}{3 y})}^{n}$ $= \frac{1}{z} \sum_{n = 0}^{\infty} \frac{2^{n}}{z^{n}} = \sum_{n = 0}^{\infty} \frac{2^{n}}{z^{n + 1}} \Rightarrow g (z) = - 2 \sum_{n = 0}^{\infty} \frac{z^{n}}{3^{n + 1}} + \sum_{n = 0}^{\infty} \frac{2^{n}}{z^{n + 1}}$
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