Finding laurent series expansion at Infinity

Question

Find the laurent series expansion at $\infty$ of the follwoing function:

$\dfrac{1}{z^2-8z+25}$

Consider $\dfrac{1}{z^2-8z+25}$,Replace $z=\frac{1}{w}$,we get

$\dfrac{1}{z^2-8z+25}=\dfrac{w^2}{25(w^2-\frac{8}{25}w+\frac{1}{25})}$

$(w^2-\frac{8}{25}w+\frac{1}{25})=(w-\frac{4}{25})^2+(\frac{3}{25})^2$

Hence $\dfrac{w^2}{25(w^2-\frac{8}{25}w+\frac{1}{25})}$

$=\dfrac{w^2}{25((\frac{3}{25})^2\{1+\dfrac{(w-\frac{4}{25})^2}{\frac{3}{25}}\})} $

The next step involves just the expansion of $\{1+\dfrac{(w-\frac{4}{25})^2}{\frac{3}{25}})\}^{-1}$

I dont understand if its going in the right direction or not

Can someone please help if I should proceed further

Answer 1

Hint:

$\dfrac{1}{z^2-8z+25}=\dfrac{w^2}{1-(8w-25w^2)}$

Now use

$\dfrac{1}{(1-a)}=1+a+a^2....$ formula above

Answer 2

Alternatively, notice that \begin{align} \frac{1}{z^2-8z+25} &= \frac{1}{(z-4+3i)(z-4-3i)} = \frac{1}{6i}\left(\frac{1}{z-4-3i}-\frac{1}{z-4+3i}\right)\\ &= \frac{1}{6i}\left(\frac{w}{1-w(4+3i)}-\frac{w}{1-w(4-3i)}\right), \end{align} and then use the geometric series formula on each of the terms.

Answer 3

No, you are not in the right direction. You should perform a partial fractions decomposition:$$\frac{w^2}{25\left(w^2-\frac8{25}w+\frac1{25}\right)}=\frac{\frac i6}{\frac{4+3i}{25}-w}-\frac{\frac i6}{\frac{4-3i}{25}-w}$$and then apply twice the equality $\frac1{1-z}=1+z+z^2+z^2+\cdots$