2
$\begingroup$

How would I find the residue at $z_0=0$ of $$f(z)=\frac{\sinh(z)}{z^4(1-z^2)}$$I tried writing it as a series and reach $$\frac{1}{1-z^2}\sum_{n=0}^\infty \frac{z^{2n-3}}{(2n+1)!}$$ and then don't know where to go form there. Any help/hints appreciated.

$\endgroup$
2
$\begingroup$

$$\sinh z = z + \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots$$

$$\frac{1}{z^4}\sinh z = \frac{1}{z^3} + \frac{1}{z3!} + \frac{z}{5!} + \cdots$$

Now,

$$\frac{1}{1-z^2} = 1 + z^2 + z^4 + \cdots$$

So then, multiplying the series (note we only want to the the $z^{-1}$ terms, so no need to multiply every term)

$$\frac{1}{z^4(1-z^2)}\sinh z = \\ \left(1 + z^2 + z^4 + \cdots\right)\left(\frac{1}{z^3} + \frac{1}{z3!} + \frac{z}{5!} + \cdots\right) = \\ \cdots + \frac{1}{z3!} + \frac{z^2}{z^3} + \cdots =\\ \cdots + \frac{7}{6z} \cdots $$

So the residue is $\frac{7}{6}$

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

I don't know why you need an infinite series. Write your function as

$$f(z) = \frac{(\sinh{z})/z}{z^3 (1-z^2)}$$

The numerator is analytic, while the denominator has poles at $z=0, \pm 1$. The residue at $z=0$ is

$$\frac{1}{2} \lim_{z \rightarrow 0} \frac{d^2}{dz^2}[z^3 f(z)] = \frac{1}{2} \frac{d^2}{dz^2} \left[\frac{(\sinh{z})/z}{1-z^2} \right ]_{z=0}$$

The algebra is pretty terrible, so I leave it to you; the answer is:

$$\text{Res}_{z=0} f(z) = \frac{7}{6}$$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Hint:
$f(z)$ has a pole of the order $3$ (why?) at $z_0=0.$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.