# Inverz Z-transform of square root with branch cut

I would need some guidance in the below. I am more or less confident about what I am doing but I still have a couple of questions that I cannot confidently answer.

Consider a sequence $x_n$ of real numbers for $n=0,1,2,\dots$ (causal sequence). In my problem $x_n$ are defined by a non-linear recursion equation which can be Z-transformed yielding an exact solution for the transform $$X(s)\equiv\sum_{n=0}^{\infty}x_ns^{-n}= \sqrt{\frac{s^2}{(s-e^{i\phi})(s-e^{-i\phi})}},$$ where $s$ is a complex variable and $\phi$ is real. From this we obtain the terms of the sequence as $$x_n=\oint_C\frac{ds}{2\pi i}s^{n-1} \sqrt{\frac{s^2}{(s-e^{i\phi})(s-e^{-i\phi})}},\qquad n=0,1,\dots,$$ where $C$ is a closed contour that (i) goes around the origin in a counterclockwise sense, (ii) is large enough to be fully contained in the region of convergence (ROC) of $X(s)$, and (iii) should contain all the poles of $X(s)$ (and perhaps all other non-analytic/singular points?).

Questions:

• from complex integration theory and the definition/representation of $X$ as a power series in $1/s$ it is clear that $C$ should contain all the poles. But what if - as in my case - there are other singular points or cuts in the finite complex plane? Does the inverse Z-transform apply here too?
• $X(s)$ has two branch points $s=e^{\pm i\phi}$ with $1/\sqrt{s}$ (integrable) singularity so the integral above is surely convergent for all $n$. (Note that $z=\infty$ is not a branch point, neither is $s=0$.) If I take the principal branch of the square root (with its cut on the negative real line) I get the correct results for $x_n$ (checked with Mathematica and numerical integration). But can I take a different branch of the square root?
• Suppose I take the other commonly used branch of the square root which has a cut on the positive reals. Then the cut of $X(s)$ would extend to infinity (although $s=\infty$ is not a branch point) and hence no matter how $C$ is chosen it would not be fully contained in the region of analyticity. How can we resolve this?
• Can in general $x_n$ depend on the choice of the branch of the square root? (It should not; the recursion gives unique solution...)

Many thanks

$\frac{s^2}{(s-e^{i \phi})(s-e^{-i \phi})}$ has two zeros/poles of odd order : at $s = e^{\pm i \phi}$. Thus $X(s) = (\frac{s^2}{(s-e^{i \phi})(s-e^{-i \phi})})^{1/2}$ has two branch points at $s = e^{\pm i \phi}$. If you put a branch cut on the line segment $[e^{i \phi} , e^{-i\phi}]$ then it is analytic on $\mathbb{C} \setminus [e^{i \phi}, e^{-i\phi}]$. Therefore, is has a Laurent series $$X(s) = \sum_{n=\color{red}{-\infty}}^\infty x_n s^n \quad ( |s| > 1), \qquad x_n=\frac{1}{n! 2\pi i }\int_{|s| = r} \frac{X(s)}{s^{n+1}}ds\quad (r > 1)$$
Also, since $X(1/s)$ is analytic at $s=0$, then $x_n = 0$ for $n > 0$, and hence $$X(s) = \sum_{n=-\infty}^0 x_n s^n \quad ( |s| > 1)$$
Finally for $Re(s) > 0,\ \displaystyle\int_0^\infty t^{n-1} e^{-st}dt = s^{-n} (n-1)!$ which means that $$X(s)-x_0 =\int_0^\infty f(t) e^{-st}dt \quad ( Re(s) > 1), \qquad f(t) = \sum_{n=1}^\infty \frac{x_n}{(n-1)!} t^{n-1}$$ where the series for $f(t)$ converges because..
• I agree with what you write (except that the $n!$ should not be there in your formula for $x_n$) but where are you trying to get at? – Andras Vanyolos Apr 21 '17 at 20:33
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