Expand in Laurent series $f(z) = \frac{(z-2)}{ z^2 + 3z - 4}$ , by powers of $z$

Question

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Hello , i don't understand how to solve this function. I just don't get it. If someone has more experience with Laurent series , I would be grateful to get an answer.

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

function

what i've got so far , but i don't think that is correct

UPDATE #1:

Well , wrote the function as $f(z) = \frac{A}{(z - 1)} + \frac{B}{(z + 4)}$

then , i wrote $z-2 = (A+B)z+4A-1B$

then , i made a system of equations like

$A+B = 1$

$4A - 1B = -2$

and got that $a = \frac{-1}{5}$ and $b =\frac{6}{5}$ , so

$f(z) = \frac{-1/5}{z - 1} + \frac{6/5}{z + 4}$

Answer 1

Sonce $0$ is not a singularity of your functions, your Laurent series is just a Taylor series. You started well:\begin{align}\frac{z-2}{z^2+3z-4}&=\frac15\times\frac1{1-z}-\frac65\times\frac1{4+z}\\&=\frac15\times\frac1{1-z}-\frac3{10}\times\frac1{1+\frac z4}\\&=\frac15\sum_{n=0}^\infty z^n-\frac3{10}\sum_{n=0}^\infty\left(-\frac14\right)^nz^n\end{align}if $|z|<1$.