I'm trying to calculate the residua of the following complex function but am encountering problems trying to determine its poles:
$$f(z)=\frac{\sin(z)}{z^4}$$
Expanding the denominator shows that we have a pole at $z=0$ of order four.
We also check the numerator for any removable singularities, and we see that:
$$\sin(z)=0, \forall z=n\pi, n \in \mathbb N$$
Now since the pole we have is $z=0$, we have a removable singularity for $z=0$ in the numerator with $\sin(0)=0$, where, in the above, $n=0$.
Does this mean that, with the one removable singularity, the order of the denominator is reduced by one, to give $z^3$ in the denominator?
If so, what formula or means should I use to calculate the residue at z=0? Since we have a removable singularity, its residue should be zero. I found online that instead of using the formula for a pole of order $m$ People evaluated the limit of f(z) directly i.e:
$$\lim_{z\to0}\frac {\sin(z)}{z^4}$$
Why does this hold instead of having to use the formula for evaluating a pole of order $m$?
Best Answer
You have a pole of order $3$. The intuitive idea is that one $z$ is "absorbed" by the $\sin z$ , the other $3$ are untouched.
You do not have a removable singularity (it is a pole!). You would have if there was a $z$ but there is not, so...
How do you make all of this formal? Just write down the Laurent series in $0$..
So you have
$$\sin z = z - \frac{z^3}6 + \frac{z^5}{5!} - \dots $$
So the Laurent expansion of $\frac{\sin z}{z^4} $ is
$$\frac{\sin z}{z^4} = \frac 1{z^3} - \frac1{6z} + \frac{z}{5!} - \dots $$
So the pole is of order $3$, and the residue is $-\frac 16$