developp at laurent series 1) f(z) =(1/(z−2)) 2)g(z) =(3/(z^2 −3z +2)) 3)h(z) =(1/(z^2 +4))


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Question Number 63560 by mathmax by abdo last updated on 05/Jul/19

$d e v e l o p p a t l a u r e n t s e r i e s$ $1) f (z) = \frac{1}{z - 2}$ $2) g (z) = \frac{3}{z^{2} - 3 z + 2}$ $3) h (z) = \frac{1}{z^{2} + 4}$
Commented by mathmax by abdo last updated on 05/Jul/19

$t h e f o r m o f L a u r e n t s e r i e i s \sum_{n = 0}^{\infty} a_{n} {(z - z_{0})}^{n} + \sum_{k = 1}^{\infty} a_{- k} {(z - z_{0})}^{- k}$ $1) w e h a v e f (z) = \frac{1}{z - 2} = - \frac{1}{2 (1 - \frac{z}{2})}$ $c a s e 1 ∣ \frac{z}{2} ∣< 1 \Rightarrow f (z) = - \frac{1}{2} \sum_{n = 0}^{\infty} {(\frac{z}{2})}^{n} = - \frac{1}{2} \sum_{n = 0}^{\infty} \frac{z^{n}}{2^{n}}$ $c a s e 2 ∣ \frac{z}{2} ∣> 1 f (z) = \frac{1}{z (1 - \frac{2}{z})} a n d ∣ \frac{2}{z} ∣< 1 \Rightarrow f (z) = \frac{1}{z} \sum_{n = 0}^{\infty} {(\frac{2}{z})}^{n}$ $= \sum_{n = 0}^{\infty} \frac{2^{n}}{z^{n + 1}} = \frac{1}{z} + \frac{2}{z^{2}} + \frac{4}{z^{3}} + . . . .$
Commented by mathmax by abdo last updated on 10/Jul/19
$2) z^2 −3z +2 =0 →Δ=9−8=1 ⇒z_1 =((3+1)/2) =2 and z_2 =((3−1)/2) =1 ⇒ g(z) =(3/((z−2)(z−1))) =−3((1/(z−1))−(1/(z−2))) if ∣z∣<1 ⇒∣z∣<2 ⇒g(z)=3((1/(1−z)) −(1/(2−z))) =3((1/(1−z)) −(1/(2(1−(z/2))))) =3{Σ_(n=0) ^∞ z^n −Σ_(n=0) ^∞ ((z/2))^n } =3Σ_(n=0) ^∞ (1−(1/2^n ))z^n if 1<∣z∣<2 ⇒∣(1/z)∣<1 and ∣(z/2)∣<1 ⇒ g(z) =3{ (1/(z((1/z)−1))) −(1/(2(1−(z/2))))}=((−3)/z) (1/(1−(1/z))) −(3/2) (1/(1−(z/2))) =((−3)/z)Σ_(n=0) ^∞ ((1/z))^n −(3/2)Σ_(n=0) ^∞ ((z/2))^n =−3Σ_(n=0) ^∞ (1/z^(n+1) ) −(3/2) Σ_(n=0) ^∞ (z^n /2^n )$
$2) z^{2} - 3 z + 2 = 0 \to Δ = 9 - 8 = 1 \Rightarrow z_{1} = \frac{3 + 1}{2} = 2 a n d z_{2} = \frac{3 - 1}{2} = 1 \Rightarrow$ $g (z) = \frac{3}{(z - 2) (z - 1)} = - 3 (\frac{1}{z - 1} - \frac{1}{z - 2})$ $i f ∣ z ∣< 1 \Rightarrow∣ z ∣< 2 \Rightarrow g (z) = 3 (\frac{1}{1 - z} - \frac{1}{2 - z})$ $= 3 (\frac{1}{1 - z} - \frac{1}{2 (1 - \frac{z}{2})}) = 3 {\sum_{n = 0}^{\infty} z^{n} - \sum_{n = 0}^{\infty} {(\frac{z}{2})}^{n}}$ $= 3 \sum_{n = 0}^{\infty} (1 - \frac{1}{2^{n}}) z^{n}$ $i f 1 <∣ z ∣< 2 \Rightarrow∣ \frac{1}{z} ∣< 1 a n d ∣ \frac{z}{2} ∣< 1 \Rightarrow$ $g (z) = 3 {\frac{1}{z (\frac{1}{z} - 1)} - \frac{1}{2 (1 - \frac{z}{2})}} = \frac{- 3}{z} \frac{1}{1 - \frac{1}{z}} - \frac{3}{2} \frac{1}{1 - \frac{z}{2}}$ $= \frac{- 3}{z} \sum_{n = 0}^{\infty} {(\frac{1}{z})}^{n} - \frac{3}{2} \sum_{n = 0}^{\infty} {(\frac{z}{2})}^{n}$ $= - 3 \sum_{n = 0}^{\infty} \frac{1}{z^{n + 1}} - \frac{3}{2} \sum_{n = 0}^{\infty} \frac{z^{n}}{2^{n}}$
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