Laurent series expansion of a complex function: $\frac{(z+1)}{z(z-4)^3}$

Question

Find the Laurent series for

$\frac{(z+1)}{z(z-4)^3} \in 0 < |z-4| < 4$.

I get you have to write the denominator in another way, but what are the intermediate steps?

Answer 1

$\frac{z+1}{z(z-4)^3}=\frac{1}{(z-4)^3}+\frac{1}{z(z-4)^3}$. For the second, you can use simple fractions (ie, write it as $\frac{A}{z}+\frac{Bz^2+Cz+D}{(z-4)^3}$

Answer 2

One may set $u=z-4$ then write $$ \begin{align} \frac{z+1}{z(z-4)^3}&=\frac{u+5}{(u+4)u^3}\\ &=\frac1{4u^3}\times(u+5)\times \frac1{1+\dfrac{u}4}\\ &=\frac1{4u^3}\times(u+5)\times \sum_{n=0}^\infty(-1)^n \dfrac{u^n}{4^n}\qquad (|u|<4)\\ &=(u+5)\times \sum_{n=0}^\infty(-1)^n \dfrac{u^{n-3}}{4^{n+1}}\\ &=\frac5{4u^3}-\frac1{16u^2}+\frac1{64u}-\sum_{n=0}^\infty(-1)^n \dfrac{u^n}{4^{n+4}} \end{align} $$ Finally, for $0<|z-4|<4$, one has the Laurent series expansion

$$ \frac{z+1}{z(z-4)^3}=\frac5{4(z-4)^3}-\frac1{16(z-4)^2}+\frac1{64(z-4)}-\sum_{n=0}^\infty(-1)^n \dfrac{(z-4)^n}{4^{n+4}}. $$