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

$$findlaurantserieslf$$$$$$$$\left(1\right)f\left(z\right)=\frac{1}{z-1}+\frac{1}{z+2},\mid z\mid >1$$$$$$$$\left(2\right)f\left(z\right)=\frac{1}{z-2}-\frac{2}{z-3},\mid z\mid <3$$

Answered by mathmax by abdo last updated on 17/Jul/21

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

Answered by Olaf_Thorendsen last updated on 17/Jul/21

$$\left(1\right)$$$$f\left(z\right)=\frac{1}{z-1}+\frac{1}{z+2},\mid z\mid >1$$$$f\left(z\right)=\frac{1}{z}.\frac{1}{1-\frac{1}{z}}+\frac{1}{2}.\frac{1}{1+\frac{z}{2}}$$$$\mid z\mid >1\Rightarrow \mid \frac{1}{z}\mid <1$$$$f\left(z\right)=\frac{1}{z}.\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{z^n}+\frac{1}{2}.\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{z^n}{2^n}$$$$f\left(z\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{z^n}+\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{z^n}{2^{n+1}}$$$$\left(2\right)$$$$f\left(z\right)=\frac{1}{z-2}-\frac{2}{z-3},\mid z\mid <3$$$$-\frac{2}{z-3}=\frac{2}{3}.\frac{1}{1-\frac{z}{3}}=-\frac{2}{3}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{z^n}{3^n}$$$$1\mathrm{st}\mathrm{case}:\mid z\mid <2$$$$\frac{1}{z-2}=-\frac{1}{2}.\frac{1}{1-\frac{z}{2}}=-\frac{1}{2}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{z^n}{2^n}$$$$f\left(z\right)=-\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{2^{n+1}}+\frac{2}{3^{n+1}})z^n$$$$$$$$2\mathrm{nd}\mathrm{case}:2<\mid z\mid <3\Rightarrow \frac{2}{\mid z\mid }<1$$$$\frac{1}{z-2}=\frac{1}{z}.\frac{1}{1-\frac{2}{z}}=\frac{1}{z}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{2^n}{z^n}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2^{n-1}}{z^n}$$$$f\left(z\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2^{n-1}}{z^n}-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{2}{3^{n+1}}{z}^n$$

Commented by Mrsof last updated on 17/Jul/21

$$sirwhyyougive\mid z\mid <2?$$

Commented by Olaf_Thorendsen last updated on 17/Jul/21

$$becauseinthefirstcasetheserie$$$$isconvergentonlyfor\mid z\mid <2$$

Commented by Mrsof last updated on 18/Jul/21

$$thankyousir$$

Answered by mathmax by abdo last updated on 17/Jul/21

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

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