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), $\implies x(1)=0$


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

1 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 $n\ge 0$.

  • $\begingroup$ 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) ? $\endgroup$
    – Suresh
    Commented Dec 30, 2020 at 11:49
  • 1
    $\begingroup$ @Suresh: Yes, these are two out of infinitely many options. $\endgroup$
    – Matt L.
    Commented Dec 30, 2020 at 11:54

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