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When we take ztransform of unit step it is z/[z-1] And our ROC is |z|>1 But if some how the minus sign between z and 1 changes to +, will our ROC still be same as old ( |z| >1 ) or will it be opposite?

Secondly the ztransform and ROC of u(-n) and u(n) are same or not?

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    $\begingroup$ You haven't accepted any answers to your past questions. This might discourage users from answering your current and future questions. Have you never received satisfactory answers? If so, please leave comments asking for clarification. Otherwise, you might want to consider accepting the most useful answer. $\endgroup$ – Matt L. Dec 25 '19 at 18:02
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You need to understand that a function such as $F(z)=\frac{z}{z+1}$ doesn't have a region of convergence (ROC). Only an infinite series has a ROC. E.g., the $\mathcal{Z}$-transform of a sequence $x[n]$ can be expressed by the series

$$\sum_{n=-\infty}^{\infty}x[n]z^{-n}\tag{1}$$

and this expression only makes sense if the series converges for certain values of $z$. The ROC is the region of the $z$-plane for which the series $(1)$ converges.

If you turn the problem around and if you're given a function of $z$ like $F(z)$ defined above, you can ask yourself if $F(z)$ is a valid $\mathcal{Z}$-transform of some sequence $f[n]$. It turns out that $F(z)$ is the $\mathcal{Z}$-transform of two different sequences, but with different ROCs. So a $\mathcal{Z}$-transform is only completely defined by the expression for the function $and$ the corresponding ROC.

Coming back to the function $F(z)$, it's straightforward to show that it is the $\mathcal{Z}$-transform of the sequence $f_1[n]=(-1)^nu[n]$ with the ROC $|z|>1$, and of the sequence $f_2[n]=-(-1)^nu[-n-1]$ with the ROC $|z|<1$.

Concerning your last question, as mentioned above, the expression for the $\mathcal{Z}$-transform including the ROC is unique, so the two sequences $u[n]$ and $u[-n]$ cannot have the same $\mathcal{Z}$-transform and ROC. It's actually pretty easy to come up with the $\mathcal{Z}$-transform of $u[-n]$:

$$\begin{align}\mathcal{Z}\big\{u[-n]\big\}&=\sum_{n=-\infty}^0z^{-n}\\&=\sum_{n=0}^{\infty}z^n\\&=\frac{1}{1-z},\qquad |z|<1\end{align}$$

which is of course different from the $\mathcal{Z}$-transform of $u[n]$. Note that the algebraic expression as well as the ROC are different.

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