I am trying to find a function whose Laplace transform is $$\mathcal{L}(f)(z)=\frac{z}{z^3-1}.$$
I first simplify$\mathcal{L}(f)(z)$ by solving for the cube roots of unity. Hence, $$\mathcal{L}(f)(z)=\frac{z}{(z-1)(z-w)(z-\overline{w})}, \ \ \text{where} \ \ w=-\frac{1}{2}+i\frac{\sqrt{3}}{2}.$$
Using a inversion formula, $$f(t)=\sum_{j=1}^{3}\text{Res}(\mathcal{L}(f)(z)e^{zt},z=a_j),$$ I compute $$\text{Res}(\mathcal{L}(f)(z)e^{zt},w)=\lim_{z\to w}\frac{e^{zt}}{(z-1)(z-\overline{w})}=\frac{e^{wt}}{(w-1)(w-\overline{w})}.$$ My question is, how do I simplify this? The result I have states $$\text{Res}(\mathcal{L}(f)(z)e^{zt},w)=\frac{e^{wt}w}{3w^2}=\frac{\overline{w}}{3}e^{wt}.$$ Expanding out the brackets seems too tedious. For instance, if it is given $$\mathcal{L}(g)(z)=\frac{1}{z^4+1},$$ it would take too long to simplify the expression for each residue. Is there another way?
I like your method but in this case I prefer to choose partial fraction decomposition \begin{align} \dfrac{z}{z^3-1} &= \dfrac13\dfrac{1}{z-1}-\dfrac13\dfrac{z+\frac12}{z^2+z+1}-\dfrac13\dfrac{-\frac32}{z^2+z+1} \\ &= \dfrac13\dfrac{1}{z-1}-\dfrac13\dfrac{z+\frac12}{(z+\frac12)^2+\frac34}-\dfrac13\dfrac{-\frac32}{(z+\frac12)^2+\frac34} \\ &= \dfrac13\dfrac{1}{z-1}-\dfrac13\dfrac{(z+\frac12)}{(z+\frac12)^2+(\frac{\sqrt{3}}{2})^2}+\dfrac{1}{\sqrt{3}}\dfrac{(\frac{\sqrt{3}}{2})}{(z+\frac12)^2+(\frac{\sqrt{3}}{2})^2} \end{align} then $${\cal L}^{-1}\left(\dfrac{s-c}{(s-c)^2+a^2}\right)=e^{ct}\cos at$$ $${\cal L}^{-1}\left(\dfrac{a}{(s-c)^2+a^2}\right)=e^{ct}\sin at$$ with $c=-\frac12$ and $a=\frac{\sqrt{3}}{2}$, then $${\cal L}^{-1}\left(\dfrac{z}{z^3-1}\right)=\dfrac{1}{3}e^t-\dfrac{1}{3}e^{-t/2}\cos\dfrac{\sqrt{3}}{2}t\ +\dfrac{\sqrt{3}}{3}e^{-t/2}\sin\dfrac{\sqrt{3}}{2}t$$