# Complex Numbers and polar form

I am given the following information:

$$x[n]= s^n,\qquad n=0,\pm 1,\pm 2,\ldots$$ where $s=\sigma + j\omega = re^{i\theta}$ is a complex number in general.

I was wondering how the following is concluded (proof?):

For $\sigma = 0$ then $x[n] = r^n$

-
Do you really mean $\sigma=0$, and not $\theta=0$? –  Arturo Magidin Apr 30 '12 at 2:58
yes, the question was formatted/edited correctly –  rrazd Apr 30 '12 at 3:01
If $s=r e^{i \theta}$, then $s^n = r^n e^{i n \theta}$. The only case where $s^n=r^n$ is if $\theta = 2 \pi k$, where $k=0,1,2,\dots$. –  Pedro Tamaroff Apr 30 '12 at 3:03

Note that if $u=re^{i\theta}$ is a complex number, then $$u^n = (re^{i\theta})^n = r^n(e^{i\theta})^n = r^n e^{in\theta}.$$ So if $\theta=0$, then $u^n = r^n$. (This is sometimes known as DeMoivre's Forumla)
In particular, if $s = re^{i\theta}$ and $\theta=0$, then $s^n = r^n$.
Added. If $\sigma=0$, then $s$ is purely imaginary, $r=|j|$ and $\theta=\pi/2$ if $j\gt0$ and $\theta=-\pi/2$ if $\lt 0$. If $n$ is a multiple of $4$, then $s^n = r^n$; if $n=4k+2$, then $s^n=-r^n$; if $n=4k+1$, then $s^n=\mathrm{sgn}(j)ir^n$, and if $n=4k+3$ then $s^n=-\mathrm{sgn}(j)ir^n$.
@rrazd: It is true if $n$ is a multiple of $4$. –  Arturo Magidin Apr 30 '12 at 15:21