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Obtain all Tayiors & Laurents series expansions of function $\frac{(z-2)(z+2)}{(z+1)(z+4)}$ about z = 0
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Let $f(z) = \frac{(z^2-4)}{(z^2+5z+4)}$ , since the power of numerator & denominator are same ,first we devide numerator by denominator f(z) can be written as

$f(z)= 1 - [\frac{(5z+8)}{(z^2+5z+4)}] = 1 – [\frac{(5z+8)}{(z+1)(z+4)}]$

After doing partial fraction of $\frac{(5z+8)}{(z+1)(z+4)}$ ,f(z) can be written as:

$f(z) =1 - \frac{1}{z+1} - \frac{4}{z+4}$, we have to expand about z=0 therefore three cases arise

Case (i) |z|<1 ∴|z|<4

$f(z) = 1 - (1+z)^{-1} – (1+\frac{z}{4})^{-1}$

= $1 – ( 1- z + z^2 –z^3 +z^4 …….) – [1- (\frac{z}{4}) +(\frac{z}{4})^2 - (\frac{z}{4})^3…….]$

Case(ii) 1< |z|<4

$f(z)= 1 - \frac{1}{z} (1+\frac{1}{z})^{-1} \, – (1+\frac{z}{4})^{-1}$

= $1 - \frac{1}{z} [1 - (\frac{1}{z}) +(\frac{1}{z})^2-(\frac{1}{z})^3 ...] - [1- (\frac{z}{4}) +(\frac{z}{4})^2 - (\frac{z}{4})^3….]$

Case(iii) |z|>4 ∴|z|>1

$f(z) = 1 - \frac{1}{z} (1+\frac{1}{z})^{-1} \, - \frac{1}{z} (1+\frac{4}{z})^{-1}$

= $1– \frac{1}{z}[1 - (\frac{1}{z}) +(\frac{1}{z})^2-(\frac{1}{z})^3 …..]- \frac{1}{z}[1- (\frac{4}{z}) +(\frac{4}{z})^2-(\frac{4}{z})^3 ….]$