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

Question

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 $eq(2)$ ?

Answer 1

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 $n\ge 0$.