developp-at-laurent-series-1-f-z-1-z-2-2-g-z-3-z-2-3z-2-3-h-z-1-z-2-4-

Question Number 63560 by mathmax by abdo last updated on 05/Jul/19

d e v e l o p p a t l a u r e n t s e r i e s

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

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

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

Commented by mathmax by abdo last updated on 05/Jul/19

t h e f o r m o f L a u r e n t s e r i e i s \sum_{n = 0}^{\infty} a_{n} {(z - z_{0})}^{n} + \sum_{k = 1}^{\infty} a_{- k} {(z - z_{0})}^{- k}

1) w e h a v e f (z) = \frac{1}{z - 2} = - \frac{1}{2 (1 - \frac{z}{2})}

c a s e 1 ∣ \frac{z}{2} ∣< 1 \Rightarrow f (z) = - \frac{1}{2} \sum_{n = 0}^{\infty} {(\frac{z}{2})}^{n} = - \frac{1}{2} \sum_{n = 0}^{\infty} \frac{z^{n}}{2^{n}}

c a s e 2 ∣ \frac{z}{2} ∣> 1 f (z) = \frac{1}{z (1 - \frac{2}{z})} a n d ∣ \frac{2}{z} ∣< 1 \Rightarrow f (z) = \frac{1}{z} \sum_{n = 0}^{\infty} {(\frac{2}{z})}^{n}

= \sum_{n = 0}^{\infty} \frac{2^{n}}{z^{n + 1}} = \frac{1}{z} + \frac{2}{z^{2}} + \frac{4}{z^{3}} + \dots .

Commented by mathmax by abdo last updated on 10/Jul/19

2) z^{2} - 3 z + 2 = 0 \to Δ = 9 - 8 = 1 \Rightarrow z_{1} = \frac{3 + 1}{2} = 2 a n d z_{2} = \frac{3 - 1}{2} = 1 \Rightarrow

g (z) = \frac{3}{(z - 2) (z - 1)} = - 3 (\frac{1}{z - 1} - \frac{1}{z - 2})

i f ∣ z ∣< 1 \Rightarrow∣ z ∣< 2 \Rightarrow g (z) = 3 (\frac{1}{1 - z} - \frac{1}{2 - z})

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

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

i f 1 <∣ z ∣< 2 \Rightarrow∣ \frac{1}{z} ∣< 1 a n d ∣ \frac{z}{2} ∣< 1 \Rightarrow

g (z) = 3 {\frac{1}{z (\frac{1}{z} - 1)} - \frac{1}{2 (1 - \frac{z}{2})}} = \frac{- 3}{z} \frac{1}{1 - \frac{1}{z}} - \frac{3}{2} \frac{1}{1 - \frac{z}{2}}

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

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