Find the Laurent Expansion for $$\frac{1}{z^2-3z-4}$$ which converges for $1<|z|<3$
I cant seem to crack this question so any help will be appreciated.
2026-03-29 19:11:12.1774811472
Find the Laurent Expansion for $\frac{1}{z^2-3z-4}$
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Did you realize that $z^2- 3z- 4= (z-4)(z+1)$ so that (partial fractions) $\frac{1}{z^2- 3z- 4}= \frac{A}{z- 4}+ \frac{B}{z+ 1}$? Multiplying on both sides by $z^2- 3z- 4$ we get $1= A(z+1)+ B(z- 4)$ for all z. Taking z= 4, 1= 5A so A= 1/5. Taking z= -1, 1= -5B so B= -1/5.
That is, $\frac{1}{z^2- 3z- 4}= \frac{1}{5}\frac{1}{z- 4}- \frac{1}{5}\frac{1}{z+ 1}= \frac{1}{5}\frac{-1}{1- \frac{z}{4}}- \frac{1}{5}\frac{1}{1- (-z)}$. Now, the "geometric series" $1+ r+ r^2+ \cdot\cdot\cdot= \frac{1}{1- r}$ so $\frac{1}{z^2- 3z- 4}= -\frac{1}{5}\left(1+ \frac{z}{4}+ \frac{z^2}{16}+ \cdot\cdot\cdot\right)- \frac{1}{5}\left(1- z+ z^2- \cdot\cdot\cdot\right)$. Now combine like powers.