# confusion related to finding inverse Z-Transform using Complex Integral Method

I am facing problems related to evaluation of inverse Z-Transform using Complex Integral Method;
Consider $$X(z)=\frac{z}{z-2}$$ and $$ROC: |z|>2$$
then, $$x(n)= \frac{1}{2\pi j}\oint_c X(z) z^{n-1} \, dz$$ $$\implies x(n)= \frac{1}{2\pi j}\oint_c \frac{z^n}{z-2} \, dz \quad \dots (1)$$ $$\implies x(-1)=\frac{1}{2\pi j}\oint_c \frac{1}{z(z-2)} \, dz$$ $$=z\frac{1}{z(z-2)}|_{z=0} + (z-2)\frac{1}{z(z-2)}|_{z=2}$$ $$=-\frac{1}{2} + \frac{1}{2} =0$$ which is obvious as $$x(n)=2^n u(n)$$ (by inspection), $$\implies x(-1)=0$$

But for the same $$X(z)$$ , if $$ROC: |z|<2$$
then, from $$eq(1)$$ , we get: $$x(1)=\frac{1}{2\pi j}\oint_c \frac{z}{z-2} \, dz \quad \dots (2)$$ $$=2 \quad \text{(from Cauchy's Integral Theorem)}$$ but $$x(n)=-2^n u(-n-1)$$ (by inspection), $$\implies x(1)=0$$

so,

1. How $$c$$ is defined in Complex Integral Method for a given $$ROC$$ ?
2. How can we evaluate $$eq(2)$$ ?

The closed contour $$C$$ must lie inside the region of convergence, so for the ROC $$|z|<2$$ you have no poles inside $$C$$ for $$n\ge 0$$, hence $$x[n]=0$$ for $$n\ge 0$$.

• Sir, then can we say that for $ROC : |z| > 2$ , $C$ is a circle given by the equation $|z|=2+\epsilon$ and for $ROC : |z| < 2$ , $C$ is $|z|=2-\epsilon$ (where $\epsilon$ is an infinitesimal value) ? Commented Dec 30, 2020 at 11:49
• @Suresh: Yes, these are two out of infinitely many options. Commented Dec 30, 2020 at 11:54