Higher order poles, how high?

Question

Clarification: I claim, for function $g(z)$, analytic and nonzero at $z=0$, if I have the function $f(z)=g(z)/z^n$ there is no use in trying to find poles of order smaller than n. And I would be justified in immediately multiplying $f(z)$ with $z^n$ and then start searching for the residue.

Following expression is just an example: say I have want to find the residue at $z=0$ for

$$\frac1{z^3}\frac1{\sqrt{1-z^2}}$$

I can see that the expression will not take a finite value at $z=0$ until I multiply by $z^3$. So to find the residue can I simply start with multiplying with $z^3$? If that limit ends up being $0$, does that mean that I can stop trying to find higher order poles (in other words the residue is actually $0$)?

Answer 1

The first step is to calculate the limit: Since : $lim_{z→0}(z-0)^3f(z)=1 \Rightarrow$

$0$ is a pole of order 3 of $f$.

Thus the residue of $f$ in $0$ is:

$\frac1{(3-1)!}lim_{z→0}[(z-0)^3f(z)]^{(3-1)}=\frac12lim_{z→0}[z^3\frac1{z^3}\frac1{\sqrt{1-z^2}}]^2=\frac12$

Answer 2

Multiplying by $z^3$ won't help. If the limit after that multiplication was $0$, it would only mean you multipied by a too high power of $z$, i.e. the pole order is lower than $3$, but the residue can't be deduced from that.

The function $\displaystyle f(z) = \frac{1}{z^3} \cdot \frac{1}{\sqrt{1-z^2}}$ can be expanded into Laurent series

$$f(z) = \sum_{n=-\infty}^{\infty} a_n \cdot z^n$$

and $\mathrm{res}_{z=0} \, f(z) = a_{-1}$. The first step is to find that expansion. As you may know,

$$(1-w)^{\mu} = \sum_{n=0}^{\infty} \binom{\mu}{n} \cdot w^n,$$

hence for $w = z^2$ and $\mu = -\frac{1}{2}$ we obtain

$$\frac{1}{\sqrt{1-z^2}} = \sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n} \cdot z^{2n} = 1 -\frac{1}{2}z^2 + \ldots.$$

Multiplynig by $\frac{1}{z^3}$ only shifts the series, so

$$f(z) = \sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n} \cdot z^{2n-3} = \frac{1}{z^3} - \frac{1}{2} \cdot \frac{1}{z} + \ldots$$

Thus $a_{-1} = -\frac{1}{2}$.

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In general you start with

$$f(z) = \sum_{n=-\infty}^{\infty} a_n \cdot z^n.$$

If you check that $\displaystyle \lim_{z \to 0} z^k \cdot f(z) = g \neq \infty$, it only means that actually

$$f(z) = \sum_{n=-k}^{\infty} a_n \cdot z^n$$

and $a_{-k} = g$. Still doesn't allow to conclude what $a_{-1}$ is without determining the expansion.

Answer 3

Recall that the residue of a complex function $f(z)$ at some point $z=w$ in the complex z-plane is given by the coefficient of $\dfrac{1}{z-w}$ (namely, $c_{-1}$) in the Laurent expansion of the function about $z=w$:

$$f(z) = \displaystyle\sum_{n=-\infty}^{\infty} c_n(z-w)^n$$

In your case, $w=0$ and $f(z)=\dfrac{1}{z^3}\dfrac{1}{\sqrt{1-z^2}}$. This has Laurent expansion:

$$f(z)=\dfrac{1}{z^3}\left[1-\dfrac{z^2}{2} - \dfrac{z^4}{8} -…\right] = \dfrac{1}{z^3} - \dfrac{1}{2z} -\dfrac{z}{8} + ... $$

Clearly this has residue $-\dfrac{1}{2}$ by reading off the coefficient of $\dfrac{1}{z}$, but note that multiplying through by $z^3$ and taking the limit as $z\to 0$ doesn't extract the coefficient of $\dfrac{1}{z}$ (which is what you need).

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Instead, observe that if you multiply through by $z^3$ and then differentiate twice, you're left with:

$$ \dfrac{d^2}{dz^2}\left[1-\dfrac{1}{2}z^2-\dfrac{z^3}{8} + …\right] = 2\cdot -\dfrac{1}{2} +O(z^2)$$

So

$$-\dfrac{1}{2} = \dfrac{1}{2}\displaystyle\lim_{z\to 0}\dfrac{d^2}{dz^2}\left[z^3f(z)\right] \not= \displaystyle\lim_{z\to 0}[z^3f(z)]$$

In general.