- #1

- 525

- 16

[tex]\lim_{|z|\to\infty} \frac{X(z)}{z} = 0[/tex]

Is that sufficient to guarantee that

[tex]\frac{1}{2\pi i}\oint_C X(z) z^{n-1} dz = 0 [/tex]

for ##n < 0## (where ##C## is the unit circle)?

My thinking is that the residue at infinity is

[tex] \operatorname{Res}_{\infty}\left[X(z) z^{n-1} \right] = \lim_{|z| \to\infty} X(z) z^n = 0[/tex]

for ## n < 0 ## because of the stated property of ##X(z)##. Then, since we've specified that there are no other poles outside the unit circle, the integral would have to be proportional to this residue, and thus it would have to be zero. Is that correct?

Alternatively, suppose I know that ##X(z)## has at least as many poles as zeros, and again all the poles are inside the unit circle. Is that enough to guarantee that the above integral will be zero for ##n<0##? I know it's true for rational ##X(z)## but I'm not quite sure how to prove it in general.