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Question Number 148082 by tabata last updated on 25/Jul/21

Commented by tabata last updated on 25/Jul/21

$$helpmemsrolaf$$

Answered by Olaf_Thorendsen last updated on 25/Jul/21

$$f\left(z\right)=\frac{z+4}{z^2(z^2+3z+2)},\mid z\mid >2$$$$f\left(z\right)=\frac{z+4}{z^2(z+1)(z+2)}$$$$f\left(z\right)=\frac{2}{z^2}-\frac{5}{2z}.+\frac{3}{z+1}-\frac{1}{2(z+2)}$$$$f\left(z\right)=\frac{2}{z^2}-\frac{5}{2z}+\frac{3}{z}.\frac{1}{1+\frac{1}{z}}-\frac{1}{2z}.\frac{1}{1+\frac{2}{z}}$$$$f\left(z\right)=\frac{2}{z^2}-\frac{5}{2z}+\frac{3}{z}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{z^n}-\frac{1}{2z}.\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{2^n}{z^n}$$$$f\left(z\right)=\frac{2}{z^2}-\frac{5}{2z}+\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n(3-2^{n-1})\frac{1}{z^{n+1}}$$$$f\left(z\right)=\frac{2}{z^2}-\frac{5}{2z}-\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^n(3-2^{n-2})\frac{1}{z^n}$$$$f\left(z\right)=-\frac{1}{z^2}-\underset{n=3}{\sum ^{\mathrm{\infty }}}{(-1)}^n(3-2^{n-2})\frac{1}{z^n}$$$$(if\mid z\mid >2)$$

Commented by tabata last updated on 25/Jul/21

$$putsir\frac{3}{z+1}isconvergajust\mid z\mid <1how$$$$$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{\left(\frac{1}{z}\right)}^n$$

Commented by tabata last updated on 25/Jul/21

$$ithink=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{\left(z\right)}^n$$

Commented by qaz last updated on 25/Jul/21

$$\mid \mathrm{z}\mid >2\Rightarrow \frac{1}{\mid \mathrm{z}\mid }<\frac{2}{\mid \mathrm{z}\mid }<1$$

Commented by mathmax by abdo last updated on 25/Jul/21

$$\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{z}+4}{{\mathrm{z}}^2({\mathrm{z}}^2+3\mathrm{z}+2)}=\frac{\mathrm{z}+4}{{\mathrm{z}}^2}(\frac{1}{\mathrm{z}+1}-\frac{1}{\mathrm{z}+2})$$$$=\frac{1}{{\mathrm{z}}^2}(\frac{\mathrm{z}+4}{\mathrm{z}+1}-\frac{\mathrm{z}+4}{\mathrm{z}+2})=\frac{1}{{\mathrm{z}}^2}(1+\frac{3}{\mathrm{z}+1}-1-\frac{2}{\mathrm{z}+2})$$$$=\frac{3}{{\mathrm{z}}^2(\mathrm{z}+1)}-\frac{2}{{\mathrm{z}}^2(\mathrm{z}+2)}\mathrm{we}\mathrm{have}\mid \mathrm{z}\mid >2\Rightarrow \mid \frac{1}{\mathrm{z}}\mid <\frac{1}{2}\mathrm{and}\mid \frac{2}{\mathrm{z}}\mid <1$$$$\mathrm{f}\left(\mathrm{z}\right)=\frac{3}{{\mathrm{z}}^3(1+\frac{1}{\mathrm{z}})}-\frac{2}{{\mathrm{z}}^3(1+\frac{2}{\mathrm{z}})}$$$$=\frac{3}{{\mathrm{z}}^3}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{z}}^{\mathrm{n}}}-\frac{2}{{\mathrm{z}}^3}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\left(\frac{2}{\mathrm{z}}\right)}^{\mathrm{n}}$$$$=3\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{z}}^{\mathrm{n}+3}}-2\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-2)}^{\mathrm{n}}}{{\mathrm{z}}^{\mathrm{n}+3}}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}+{(-2)}^{\mathrm{n}+1})\times \frac{1}{{\mathrm{z}}^{\mathrm{n}+3}}$$

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