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Question Number 147623 by mari last updated on 22/Jul/21

Commented by tabata last updated on 22/Jul/21

$$\left(2\right)itisclearleythatf\left(z\right)=\frac{1}{z-1}-\frac{1}{z-2}$$$$$$$$\left(a\right)\mid z\mid <1$$$$$$$$f\left(z\right)=-\frac{1}{1-z}+\frac{1}{2}.\frac{1}{(1-\frac{z}{2})}=-\underset{n=0}{\sum ^{\mathrm{\infty }}}{z}^n+\frac{1}{2}\underset{n=0}{\sum ^{\mathrm{\infty }}}{\left(\frac{z}{2}\right)}^n=-\underset{n=0}{\sum ^{\mathrm{\infty }}}{z}^n+\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{z^n}{2^{2+1}}$$$$$$$$\left(b\right)1<\mid z\mid <2$$$$$$$$f\left(z\right)=\frac{1}{z}.\frac{1}{(1-\frac{1}{z})}+\frac{1}{2}.\frac{1}{(1-\frac{z}{2})}=\frac{1}{z}\underset{n=0}{\sum ^{\mathrm{\infty }}}{\left(\frac{1}{z}\right)}^n+\frac{1}{2}\underset{n=0}{\sum ^{\mathrm{\infty }}}{\left(\frac{z}{2}\right)}^n=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{z^{n+1}}+\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{z^n}{2^{n+1}}$$$$$$$$\left(c\right)\mid z\mid >2$$$$$$$$case1:$$$$$$$$f\left(z\right)=-\frac{1}{z}.\frac{1}{(1-\frac{2}{z})}=-\frac{1}{z}\underset{n=0}{\sum ^{\mathrm{\infty }}}{\left(\frac{2}{z}\right)}^n=-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{2^n}{z^{n+1}}$$$$$$$$case2:\mid z\mid <1$$$$$$$$f\left(z\right)=-\frac{1}{1-z}=-\underset{n=0}{\sum ^{\mathrm{\infty }}}{z}^n$$$$$$$$\therefore f\left(z\right)=-\underset{n=0}{\sum ^{\mathrm{\infty }}}{z}^n-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{2^n}{z^{n+1}}$$$$$$$$⟨MT⟩$$$$$$

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