# Why there is Difference between shapes of ROC of z domain and s domain?

ROC(region of convergence) of Z domain is shown by a circular region while ROC in S domain is shown by a rectangular(approximately looking like rectangle) region

What is the reason of this difference in shapes of ROC regions?

Because, the region of convergence in the Laplace transform $$X(s) = \int_{-\infty}^{\infty} x(t) e^{-st} dt$$ is related to the weighting provided by the real part of the complex $$s = \sigma + j \omega$$; as this will yield the weight $$|e^{-st}| = e^{-\sigma t}$$ applied on the input signal $$x(t)$$, and is a function of $$\sigma$$ alone and is a rectangular (planar) region on the s-plane.
But the region of convergence in the Z-transform $$X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n}$$ is related to the weighting provided by the magnitude of the complex $$z = \sigma + j \omega$$ as given by $$|z|^{-n} = |\sigma + j\omega|^{-n} = |z|^{-n} |e^{-j n \angle{z}}| = |z|^{-n}$$, which is a circular region on th z-plane, due to the magnitude of $$z$$ being involved.