# ROC of $\mathcal{Z}$-Transform and zeros

Theorem: Let $$f(z) = \sum_{n=0}^{+\infty}a_nz^n$$ where $$z\in\mathbb{C}$$. If $$f(z_0)$$ exists for some $$z_0\in\mathbb{C}$$ then it converges for all $$z\in\mathbb{C}$$ such that $$|z|\lt|z_0|$$. Proof: It follows from the hypothesis that $$\exists M\ge0 : |a_nz_0^n|\le M$$ for all $$n\in\mathbb{N}$$. We have $$|a_nz^n|=|a_nz_0^n||\frac{z}{z_0}|^n\le M|\frac{z}{z_0}|^n$$. By the comparison theorem and the behavior of geometric series, we conclude that $$f(z)$$ converges for $$|z|<|z_0|$$.

So if we choose $$z_0$$ such that $$f(z_0) = 0$$, then $$f(z)$$ converges for $$|z|\lt|z_0|$$(Because $$f(z_0) = 0$$ means that $$f(z_0)$$ exists). In other words, ROC of $$\mathcal{Z}$$-Transform depends on the location of poles as well as zeros of $$f(z)$$ but according to the literature, ROC depends only on the location of poles. So what's my mistake here? Does the location of zeros affect the ROC?

• Why do you think you get to choose $z_0$? Oct 6, 2021 at 11:28
• @Jazzmaniac: I think the OP means to say that $z_0$ simply is a zero of $X(z)$, so $X(z_0)=0$ is satisfied, and, hence - at least according to the OP - $X(z)$ should converge for any $z$ satisfying $|z|<|z_0|$ (because it apparently converges for $z=z_0$). Oct 6, 2021 at 11:44
• I'm confused. As far as I understand poles and zeros are defined for systems but the Z-transform is applied to signals. What's the pole of a signal ? Oct 6, 2021 at 12:29
• @Hilmar: $x[n]$ is just a sequence; its interpretation is irrelevant here, isn't it? We usually compute Z-transforms of signals and of impulse responses, so both interpretations are fine I guess. Oct 6, 2021 at 13:40
• @Jazzmaniac: I'm not sure I understand it, but that's my interpretation of the question. I think the OP's argument goes as follows: if $X(z)$ converges for some $z_0$ then it must converge for all $z$ satisfying $|z|<|z_0|$. If $z_0$ is a zero of $X(z)$ then $X(z_0)=0$ holds, and, consequently, $X(z)$ converges for $z_0$, hence it converges for all $z$ satisfying $|z|<|z_0|$. Oct 6, 2021 at 16:09

Just because $$f(z)$$ converges for $$|z|\le|z_0|$$ for your choice of $$\{ z_0 : f(z_0) = 0\}$$ doesn't imply that $$f(z)$$ can't converge for $$|z| \gt |z_0|$$ also... up until the pole of next greatest magnitude.