Can any one teach me how to solve C2.(a) and (b) step by step?
C2. (a) Resolve $\frac{1}{s^2(s^2+s+1)}$ into partial fractions of the form $\frac{A}{s}+\frac{B}{s^2}+\frac{Cs+D}{s^2+s+1}$. Hence, resolve $\frac{1}{s(s^2+s+1)}$ into its partial fractions.
C2. (b) Find $$\mathcal{L}^{-1}\left\{\frac{s}{s^2+s+1}\right\}.$$
(a) The function $$ f(z)=\frac{1}{z^2(z^2+z+1)}$$ has a double pole in $z=0$ and a simple pole in $z=e^{\pm 2\pi i/3}$, since $z^2+z+1=\frac{z^3-1}{z-1}$. This gives: $$ f(z)=\frac{A}{z^2}+\frac{B}{z}+\frac{C}{z-\omega}+\frac{D}{z-\omega^2}+g(z)$$ where $g(z)$ is a holomorphic function. By computing: $$\lim_{z\to 0}z^2 f(z)=1,\quad \lim_{z\to\omega}f(z)(z-\omega)=\frac{1}{\sqrt{3}}e^{i\pi/6},\quad \lim_{z\to\omega}f(z)(z-\omega^2)=\frac{1}{\sqrt{3}}e^{-i\pi/6}$$ we have: $$ f(z)=\frac{1}{z^2}+\frac{z}{z^2+z+1}+\frac{B}{z}+g(z) $$ where $B=-1$ since the sum of the residues must be zero.
Rearranging and checking that $g(z)=0$, we get: $$ f(z) = \frac{1}{z^2}-\frac{1}{z}+\frac{e^{i\pi/6}}{\sqrt{3}(z-\omega)}+\frac{e^{-i\pi/6}}{\sqrt{3}(z-\omega^2)}=\frac{1}{z^2}-\frac{1}{z}+\frac{z}{z^2+z+1}.$$ Multypling by $z$ and rearranging we have: $$ \frac{1}{z(z^2+z+1)}=\frac{1}{z}+\frac{-3+i\sqrt{3}}{6(z-\omega)}+\frac{-3-i\sqrt{3}}{6(z-\omega^2)}=\frac{1}{z}-\frac{z+1}{z^2+z+1}.$$
(b) Since $$\frac{z}{z^2+z+1}=\frac{e^{i\pi/6}}{\sqrt{3}(z-\omega)}+\frac{e^{-i\pi/6}}{\sqrt{3}(z+\omega)}$$ and $$\mathcal{L}^{-1}\left(\frac{1}{z-\xi}\right)=e^{s\xi},$$ by linearity it follows that $$\color{red}{\mathcal{L}^{-1}\left(\frac{z}{z^2+z+1}\right)} = \frac{1}{\sqrt{3}}\left(e^{i\pi/6}e^{\omega s}+e^{-i\pi/6}e^{\omega^2 s}\right)=\color{red}{e^{-s/2}\left(\cos\frac{\sqrt{3}\,s}{2}-\frac{1}{\sqrt{3}}\sin\frac{\sqrt{3}\,s}{2}\right)}$$