find the power series exponition of f(z)=((2z+1)/(z^2 −3z+2)) about z


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Question Number 196565 by mokys last updated on 27/Aug/23

$f i n d t h e p o w e r s e r i e s e x p o n i t i o n o f$ $f (z) = \frac{2 z + 1}{z^{2} - 3 z + 2} a b o u t z_{o} = i$
	Answered by aleks041103 last updated on 27/Aug/23
	$((2z+1)/(z^2 −3z+2))=((2z+1)/((z−2)(z−1)))=(A/(z−1))+(B/(z−2)) ⇒A(z−2)+B(z−1)=2z+1 ⇒A+B=2 & 2A+B=−1 ⇒A=−3,B=5 ⇒f(z)=(5/(z−2))−(3/(z−1)) z=w+z_0 =w+i ⇒f=(5/(w−(2−i)))−(3/(w−(1−i))) to expand in z around z_0 is to expand in w around 0. −−−−−−−−−−−−−−−−−−−− Examine: Expand (1/(w−s)), w,s∈C 1) for ∣w∣<∣s∣: (1/(w−s))=−(1/s) (1/(1−(w/s)))=−(1/s)(1+(w/s)+((w/s))^2 +...) ⇒(1/(w−s))=Σ_(k=0) ^∞ (−(1/s^(k+1) ))w^k , ∣w∣<∣s∣ 2) for ∣w∣>∣s∣: (1/(w−s))=(1/w) (1/(1−(s/w)))=(1/w)(1+(s/w)+((s/w))^2 +...) ⇒(1/(w−s))=Σ_(k=1) ^∞ (s^(k−1) )w^(−k) , ∣w∣>∣s∣ −−−−−−−−−−−−−−−−−−−− Now let ∣s∣>∣v∣ and then: (a/(w−s))+(b/(w−v))= { ((Σ_(k=0) ^∞ (−(a/s^(k+1) )−(b/v^(k+1) ))w^k , ∣w∣<∣v∣<∣s∣)),((Σ_(k=1) ^∞ bv^(k−1) w^(−k) +Σ_(k=0) ^∞ (−(a/s^(k+1) ))w^k , ∣v∣<∣w∣<∣s∣)),((Σ_(k=1) ^∞ (as^(k−1) +bv^(k−1) )w^(−k) , ∣v∣<∣s∣<∣w∣)) :} Now just replace s=2−i, a=5, v=1−i, b=−3 and w=z−i and you get your answer$
	$\frac{2 z + 1}{z^{2} - 3 z + 2} = \frac{2 z + 1}{(z - 2) (z - 1)} = \frac{A}{z - 1} + \frac{B}{z - 2}$ $\Rightarrow A (z - 2) + B (z - 1) = 2 z + 1$ $\Rightarrow A + B = 2 & 2 A + B = - 1$ $\Rightarrow A = - 3, B = 5$ $\Rightarrow f (z) = \frac{5}{z - 2} - \frac{3}{z - 1}$ $z = w + z_{0} = w + i$ $\Rightarrow f = \frac{5}{w - (2 - i)} - \frac{3}{w - (1 - i)}$ $t o e x p a n d i n z a r o u n d z_{0} i s t o e x p a n d i n$ $w a r o u n d 0 .$ $- - - - - - - - - - - - - - - - - - - -$ $E x a m i n e : E x p a n d \frac{1}{w - s}, w, s \in C$ $1) f o r ∣ w ∣<∣ s ∣:$ $\frac{1}{w - s} = - \frac{1}{s} \frac{1}{1 - \frac{w}{s}} = - \frac{1}{s} (1 + \frac{w}{s} + {(\frac{w}{s})}^{2} + . . .)$ $\Rightarrow \frac{1}{w - s} = \underset{k = 0}{\sum^{\infty}} (- \frac{1}{s^{k + 1}}) w^{k}, ∣ w ∣<∣ s ∣$ $2) f o r ∣ w ∣>∣ s ∣:$ $\frac{1}{w - s} = \frac{1}{w} \frac{1}{1 - \frac{s}{w}} = \frac{1}{w} (1 + \frac{s}{w} + {(\frac{s}{w})}^{2} + . . .)$ $\Rightarrow \frac{1}{w - s} = \underset{k = 1}{\sum^{\infty}} (s^{k - 1}) w^{- k}, ∣ w ∣>∣ s ∣$ $- - - - - - - - - - - - - - - - - - - -$ $N o w l e t ∣ s ∣>∣ v ∣ a n d t h e n :$ $\frac{a}{w - s} + \frac{b}{w - v} = {\begin{cases} \underset{k = 0}{\sum^{\infty}} (- \frac{a}{s^{k + 1}} - \frac{b}{v^{k + 1}}) w^{k}, ∣ w ∣<∣ v ∣<∣ s ∣ \\ \underset{k = 1}{\sum^{\infty}} {b v}^{k - 1} w^{- k} + \underset{k = 0}{\sum^{\infty}} (- \frac{a}{s^{k + 1}}) w^{k}, ∣ v ∣<∣ w ∣<∣ s ∣ \\ \underset{k = 1}{\sum^{\infty}} ({a s}^{k - 1} + {b v}^{k - 1}) w^{- k}, ∣ v ∣<∣ s ∣<∣ w ∣ \end{cases}$ $N o w j u s t r e p l a c e$ $s = 2 - i, a = 5, v = 1 - i, b = - 3$ $a n d$ $w = z - i$ $a n d y o u g e t y o u r a n s w e r$
	Commented by mokys last updated on 27/Aug/23

	$i t h i n k v = 1 + i$
	Commented by aleks041103 last updated on 27/Aug/23

	$i n t h i s c a s e i t i s .$ $z = w + z_{0} = w + i$ $z - 1 = w + i - 1 = w - (1 - i) = w - v$ $\Rightarrow v = 1 - i$
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