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Question Number 196565 by mokys last updated on 27/Aug/23
find the power series exponition of f(z)=((2z+1)/(z^2 −3z+2)) about z_o = i
$${find}\:{the}\:{power}\:{series}\:{exponition}\:{of}\: \\ $$$${f}\left({z}\right)=\frac{\mathrm{2}{z}+\mathrm{1}}{{z}^{\mathrm{2}} −\mathrm{3}{z}+\mathrm{2}}\:\:{about}\:{z}_{{o}} \:=\:{i} \\ $$
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
$$\frac{\mathrm{2}{z}+\mathrm{1}}{{z}^{\mathrm{2}} −\mathrm{3}{z}+\mathrm{2}}=\frac{\mathrm{2}{z}+\mathrm{1}}{\left({z}−\mathrm{2}\right)\left({z}−\mathrm{1}\right)}=\frac{{A}}{{z}−\mathrm{1}}+\frac{{B}}{{z}−\mathrm{2}} \\ $$$$\Rightarrow{A}\left({z}−\mathrm{2}\right)+{B}\left({z}−\mathrm{1}\right)=\mathrm{2}{z}+\mathrm{1} \\ $$$$\Rightarrow{A}+{B}=\mathrm{2}\:\&\:\mathrm{2}{A}+{B}=−\mathrm{1} \\ $$$$\Rightarrow{A}=−\mathrm{3},{B}=\mathrm{5} \\ $$$$\Rightarrow{f}\left({z}\right)=\frac{\mathrm{5}}{{z}−\mathrm{2}}−\frac{\mathrm{3}}{{z}−\mathrm{1}} \\ $$$${z}={w}+{z}_{\mathrm{0}} ={w}+{i} \\ $$$$\Rightarrow{f}=\frac{\mathrm{5}}{{w}−\left(\mathrm{2}−{i}\right)}−\frac{\mathrm{3}}{{w}−\left(\mathrm{1}−{i}\right)} \\ $$$${to}\:{expand}\:{in}\:{z}\:{around}\:{z}_{\mathrm{0}} \:{is}\:{to}\:{expand}\:{in} \\ $$$${w}\:{around}\:\mathrm{0}. \\ $$$$−−−−−−−−−−−−−−−−−−−− \\ $$$${Examine}:\:{Expand}\:\:\frac{\mathrm{1}}{{w}−{s}},\:{w},{s}\in\mathbb{C} \\ $$$$\left.\mathrm{1}\right)\:{for}\:\mid{w}\mid<\mid{s}\mid: \\ $$$$\frac{\mathrm{1}}{{w}−{s}}=−\frac{\mathrm{1}}{{s}}\:\frac{\mathrm{1}}{\mathrm{1}−\frac{{w}}{{s}}}=−\frac{\mathrm{1}}{{s}}\left(\mathrm{1}+\frac{{w}}{{s}}+\left(\frac{{w}}{{s}}\right)^{\mathrm{2}} +…\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{{w}−{s}}=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\frac{\mathrm{1}}{{s}^{{k}+\mathrm{1}} }\right){w}^{{k}} ,\:\mid{w}\mid<\mid{s}\mid \\ $$$$\left.\mathrm{2}\right)\:{for}\:\mid{w}\mid>\mid{s}\mid: \\ $$$$\frac{\mathrm{1}}{{w}−{s}}=\frac{\mathrm{1}}{{w}}\:\frac{\mathrm{1}}{\mathrm{1}−\frac{{s}}{{w}}}=\frac{\mathrm{1}}{{w}}\left(\mathrm{1}+\frac{{s}}{{w}}+\left(\frac{{s}}{{w}}\right)^{\mathrm{2}} +…\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{{w}−{s}}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left({s}^{{k}−\mathrm{1}} \right){w}^{−{k}} ,\:\mid{w}\mid>\mid{s}\mid \\ $$$$−−−−−−−−−−−−−−−−−−−− \\ $$$${Now}\:{let}\:\mid{s}\mid>\mid{v}\mid\:{and}\:{then}: \\ $$$$\frac{{a}}{{w}−{s}}+\frac{{b}}{{w}−{v}}=\begin{cases}{\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\frac{{a}}{{s}^{{k}+\mathrm{1}} }−\frac{{b}}{{v}^{{k}+\mathrm{1}} }\right){w}^{{k}} ,\:\mid{w}\mid<\mid{v}\mid<\mid{s}\mid}\\{\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{bv}^{{k}−\mathrm{1}} {w}^{−{k}} +\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\frac{{a}}{{s}^{{k}+\mathrm{1}} }\right){w}^{{k}} ,\:\mid{v}\mid<\mid{w}\mid<\mid{s}\mid}\\{\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left({as}^{{k}−\mathrm{1}} +{bv}^{{k}−\mathrm{1}} \right){w}^{−{k}} \:,\:\mid{v}\mid<\mid{s}\mid<\mid{w}\mid}\end{cases} \\ $$$${Now}\:{just}\:{replace} \\ $$$${s}=\mathrm{2}−{i},\:{a}=\mathrm{5},\:{v}=\mathrm{1}−{i},\:{b}=−\mathrm{3} \\ $$$${and} \\ $$$${w}={z}−{i} \\ $$$${and}\:{you}\:{get}\:{your}\:{answer} \\ $$
Commented by mokys last updated on 27/Aug/23
i think v = 1+i
$${i}\:{think}\:{v}\:=\:\mathrm{1}+{i} \\ $$
Commented by aleks041103 last updated on 27/Aug/23
in this case it is. z=w+z_0 =w+i z−1=w+i−1=w−(1−i)=w−v ⇒v=1−i
$${in}\:{this}\:{case}\:{it}\:{is}. \\ $$$${z}={w}+{z}_{\mathrm{0}} ={w}+{i} \\ $$$${z}−\mathrm{1}={w}+{i}−\mathrm{1}={w}−\left(\mathrm{1}−{i}\right)={w}−{v} \\ $$$$\Rightarrow{v}=\mathrm{1}−{i} \\ $$

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