Question-148082

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

Commented by tabata last updated on 25/Jul/21

h e l p m e m s r o l a f

Answered by Olaf_Thorendsen last updated on 25/Jul/21

f (z) = \frac{z + 4}{z^{2} (z^{2} + 3 z + 2)}, ∣ z ∣ > 2

f (z) = \frac{z + 4}{z^{2} (z + 1) (z + 2)}

f (z) = \frac{2}{z^{2}} - \frac{5}{2 z} . + \frac{3}{z + 1} - \frac{1}{2 (z + 2)}

f (z) = \frac{2}{z^{2}} - \frac{5}{2 z} + \frac{3}{z} . \frac{1}{1 + \frac{1}{z}} - \frac{1}{2 z} . \frac{1}{1 + \frac{2}{z}}

f (z) = \frac{2}{z^{2}} - \frac{5}{2 z} + \frac{3}{z} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{z^{n}} - \frac{1}{2 z} . \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{2^{n}}{z^{n}}

f (z) = \frac{2}{z^{2}} - \frac{5}{2 z} + \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} (3 - 2^{n - 1}) \frac{1}{z^{n + 1}}

f (z) = \frac{2}{z^{2}} - \frac{5}{2 z} - \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} (3 - 2^{n - 2}) \frac{1}{z^{n}}

f (z) = - \frac{1}{z^{2}} - \underset{n = 3}{\sum^{\infty}} {(- 1)}^{n} (3 - 2^{n - 2}) \frac{1}{z^{n}}

(i f ∣ z ∣ > 2)

Commented by tabata last updated on 25/Jul/21

p u t s i r \frac{3}{z + 1} i s c o n v e r g a j u s t ∣ z ∣< 1 h o w

\underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} {(\frac{1}{z})}^{n}

Commented by tabata last updated on 25/Jul/21

i t h i n k = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} {(z)}^{n}

Commented by qaz last updated on 25/Jul/21

∣ z ∣> 2 \Rightarrow \frac{1}{∣ z ∣} < \frac{2}{∣ z ∣} < 1

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

f (z) = \frac{z + 4}{z^{2} (z^{2} + 3 z + 2)} = \frac{z + 4}{z^{2}} (\frac{1}{z + 1} - \frac{1}{z + 2})

= \frac{1}{z^{2}} (\frac{z + 4}{z + 1} - \frac{z + 4}{z + 2}) = \frac{1}{z^{2}} (1 + \frac{3}{z + 1} - 1 - \frac{2}{z + 2})

= \frac{3}{z^{2} (z + 1)} - \frac{2}{z^{2} (z + 2)} we have ∣ z ∣> 2 \Rightarrow∣ \frac{1}{z} ∣< \frac{1}{2} and ∣ \frac{2}{z} ∣< 1

f (z) = \frac{3}{z^{3} (1 + \frac{1}{z})} - \frac{2}{z^{3} (1 + \frac{2}{z})}

= \frac{3}{z^{3}} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{z^{n}} - \frac{2}{z^{3}} \sum_{n = 0}^{\infty} {(- 1)}^{n} {(\frac{2}{z})}^{n}

= 3 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{z^{n + 3}} - 2 \sum_{n = 0}^{\infty} \frac{{(- 2)}^{n}}{z^{n + 3}}

= \sum_{n = 0}^{\infty} {(- 1)}^{n} + {(- 2)}^{n + 1}) \times \frac{1}{z^{n + 3}}