# ROC of z-transform (derivative)

1. Oct 16, 2012

### jashua

I have a question about the z-transform:

In the (Oppenheim's Signals and Systems) book it is written:

If the z-transform of x[n] is X(z) with ROC=R, then the z-transform of nx[n] is -zdX(z)/dz with the same ROC.

I don't understand why the ROC remains unchanged. Some books say "it follows from the fact that X(z) is analytic", some other books say "note that Laurent z-transform is differentiable within the ROC". But these reasons are not clear to me at all!

Any help will be appreciated.

2. Oct 17, 2012

### jashua

I have found some other interpretations about my question:

1) Multiplication of x[n] by a linearly increasing sequence doesn't change the ROC, since we have z^{-n} factor in the z-transform.

2) There is no extra pole-zero introduced by differentiation; only their order can change.

The first one makes sense; however still it is not clear since, for example, if z=e^{jw}, then there is no decaying exponential in the z-transform.

The second one brings some other questions: Do we always have to think X(z) as a ratio of polynomials? Why don't we have extra pole-zero after differentiation?