Inverse z-transform by convolution method I've been asked to find the poles, impulse response of a system (in digital signal processing).
The transfer function was,
$$H(z) = \frac{z^2+z}{z^2-z+0.5}$$
I solved it by the partial fractions method as follows.
$$\color{blue}{H(z) = \frac{z(z+1)}{z^2-z+0.5}}$$
$$\begin{align} &\Rightarrow \frac{H(z)}{z} = \frac{z+1}{z^2-z+0.5}=\frac{z+1}{\left(z-(0.5+0.5i)\right)\left(z-(0.5-0.5i)\right)} 
\\ &\Rightarrow \frac{z+1}{\left(z-(0.5+0.5i)\right)\left(z-(0.5-0.5i)\right)} = \frac{A}{z-(0.5+0.5i)} +\frac{A^*}{z-(0.5-0.5i)}\\ & \\
& A = 0.5 - 1.5i \text{ and } A^* = 0.5+1.5i \\ &
\\& \Rightarrow H(z) = (0.5-1.5i)\frac{z}{z-(0.5+0.5i)} + (0.5+1.5i)\frac{z}{z-(0.5-0.5i)} \\&
\\& \Rightarrow \color{blue}{h(n) = \left[(0.5-1.5)(0.5+0.1i)^n +(0.5+1.5i)(0.5-0.5i)^n\right]u(n)}\end{align}$$
(This can be further reduced to terms containing $\cos(kn)$ and/or $\sin(kn)$ but that wasn't required.)

Then as a continuum, I've been asked to find the impulse response (Inverse z-transform of $H(z)$) by convolution method.
We have,
$$H(z) = \frac{z(z+1)}{z^2-z+0.5}$$
If it were of the form, $\frac{z^2}{(z-a)(z-b)}$, we can consider $F(z) = \frac{z}{z-a}$ and $G(z) =\frac{z}{z-b}$, find $f(n) = a^nu(n)$ , $g(n) = b^nu(n)$. Then the inverse could be found by convoluting $f(n)$ and $g(n)$ easily.
But in this case, the term $z+1$ in the numerator makes it difficult to rewrite $H(z)$ as a product of 2 functions of $z$ whose inverse z-transform is known.

Can somebody help in finding $2$ such functions so that it could be evaluated by convolution method ? Or is it impossible to find the inverse z-transform of this function by convolution?

I also thought of using the fact $Z(a^n \sin(\omega n)) = \dfrac{az\sin\omega}{z^2-2az\cos\omega + a^2}$, but this left me behind with $z+1$.
 A: Thanks to my professor for providing me a hint (To split $\dfrac{z+1}{z-b}$ into $\dfrac{z}{z-b}$ and $\dfrac{1}{z-b}$ and then convolute)
$$ H(z) = \dfrac{z(z+1)}{z^2-z+0.5} = \dfrac{z(z+1)}{\left(z-(0.5+0.5i)\right)\left(z-(0.5-0.5i)\right)}$$
Let $a= 0.5+0.5i$ and $b= 0.5-0.5i$.
$$H(z) = \dfrac{z}{z-a}\cdot\dfrac{z+1}{z-b}$$
Let
$$\begin{align} & H_1(z) = \dfrac{z}{z-a} \Rightarrow \color{blue}{h_1(n) = a^nu(n) }\\&H_2(z) = \dfrac{z+1}{z-b}=\dfrac{z}{z-b}+z^{-1}\cdot\dfrac{z}{z-b}\Rightarrow \color{green}{h_2(n) = b^nu(n)+b^{n-1}u(n-1)}\\&
\\& h(n) = \color{blue}{h_1(n)}*\color{green}{h_2(n)} = \sum_{k=-\infty}^{\infty}a^ku(k)b^{n-k}u(n-k) + \sum_{k=-\infty}^{\infty}a^ku(k)b^{n-k-1}u(n-k-1) 
\\&\Rightarrow h(n) =\sum_{k=0}^na^kb^{n-k} + \sum_{k=0}^{n-1}a^kb^{n-k-1} = b^n\sum_{k=0}^n\left(\frac{a}{b}\right)^k + b^{n-1}\sum_{k=0}^{n-1}\left(\frac{a}{b}\right)^k  
\\&\Rightarrow\color{red}{h(n) = \left[\frac{a^{n+1}-b^{n+1}}{a-b} +\frac{a^n - b^n}{a-b}\right]u(n) = \left[\frac{a^n(a+1)-b^n(b+1)}{a-b}\right]u(n)}\end{align} $$
Subbing $a$ and $b$,
$$\begin{align}& h(n) = \frac{(0.5+0.5i)^n(1.5+0.5i)-(0.5-0.5i)^n(1.5-0.5i)}{i}u(n) 
\\& \Rightarrow\color{blue}{h(n) = \left[(0.5+0.5i)^n(0.5-1.5i)+(0.5-0.5i)^n(0.5+1.5i)\right]u(n)}\end{align}$$
