finding the residue of the following

Question

I can find $$Res_{z=0} \frac{\sin z}{z^4} $$ but stuck with finding $$Res_{z=0} \frac{\cot z}{z^4} $$

so please help me

Answer 1

There are various ways to do this: it is a matter of judgement and experience which one of them is going to be the easiest. Here is one possibility.

Write $$f(z)=\frac{\cot z}{z^4}=\frac{f_0}{z^5}+\cdots+\frac{f_4}{z}+f_5+\cdots\ .$$ The residue we wish to compute is $f_4$. Multiplying both sides by $z^5\sin z$ and substituting known Taylor series gives $$z\Bigl(1-\frac{z^2}{2}+\frac{z^4}{24}-\cdots\Bigr) =\Bigl(z-\frac{z^3}{6}+\frac{z^5}{120}-\cdots\Bigr)(f_0+\cdots+f_4z^4+\cdots) \ .$$ Now equate coefficients of $z$ and $z^3$ and $z^5$ to give $$\displaylines{ f_0=1\cr f_2-\frac{f_0}{6}=-\frac{1}{2}\cr f_4-\frac{f_2}{6}+\frac{f_0}{120}=\frac{1}{24}\ .\cr}$$ Solving, we obtain $f_0=1$ and $f_2=-\frac{1}{3}$ and the answer: $$f_4=-\frac{1}{45}\ .$$

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An alternative would be to use the formula for the residue which gives $$\frac{1}{4!}\left.\frac{d^4}{dz^4}\Bigl(\frac{z\cos z}{\sin z}\Bigr)\right|_{z\to0}\ .$$ But to evaluate this you will have to differentiate four times: that's bad enough for a start. Then the denominator will have a zero of order $4$, so you will have to apply L'Hopital's Rule four times (though perhaps there will be shortcuts based on the well-known limit $z/\sin z\to1$ as $z\to0$). Overall, I think I would go with the above approach.

Answer 2

$\cot z = \frac{1}{\tan z}$ and $\tan z$ has a simple zero at zero.

Answer 3

In the line of @David's answer, compute the first three terms of the power series expansion of $\cot(z)$. This will give you the $z^3$ term in the numerator; it's coefficients will be our residue $a_{-1}$.

We use long division to compute the $z^3$ term of $\cot(z)$: $$ \cot(z) = \frac{\cos(z)}{\sin(z)} \\ = \frac{ 1+\frac{1}{2!}z^2 + \frac{1}{4!} z^4 + \ldots }{z + \frac{1}{3!} z^3 + \frac{1}{5!} z^5 + \ldots } \\ =\frac{1}{z} \frac{ 1+\frac{1}{2!}z^2 + \frac{1}{4!} z^4 + \ldots }{1 + \frac{1}{3!} z^2 + \frac{1}{5!} z^4 + \ldots } \\ \frac{1}{z}\bigg( 1 + \frac{1}{3}z^2 + \color{blue}{\frac{-2}{15}}z^4 + \ldots \bigg) . $$

Hence, on inspection of our original function, in this form, we can now obtain the residue.

$$ \frac{\cot(z)}{z^4} = \frac{1}{z^5} + \frac{1}{3} \frac{1}{z^3} - \color{blue}{\frac{-2}{15}} \frac{1}{z} + \ldots . $$

Note that I may have made a calculation mistake along the way, as my residue doesn't agree with @David's.