The following problems are from Brown and Churchill's Complex Variables, 8ed.

From §71 concerning Residues and Poles, problem #1d:

Determine the residue at $z = 0$ of the function $$\frac{\cot(z)}{z^4} $$

I really don't know where to start with this. I had previously tried expanding the series using the composition of the series expansions of $\cos(z)$ and $\sin(z)$ but didn't really achieve any favorable outcomes. If anyone has an idea on how I might go about solving this please let me know. For the sake of completion, the book lists the solution as $-1/45$.

From the same section, problem #1e

Determine the residue at $z = 0$ of the function $$\frac{\sinh(z)}{z^4(1-z^2)} $$

Recognizing the following expressions:

$$\sinh(z) = \sum_{n=0}^{\infty} \frac{z^{(2n+1)}}{(2n +1)!}$$ $$\frac{1}{1-z^2} = \sum_{n=0}^{\infty} (z^2)^n $$

I have expanded the series thusly:

$$\begin{aligned} \frac{\sinh(z)}{z^4(1-z^2)} &= \frac{1}{z^4} \bigg(\sum_{n=0}^{\infty} \frac{z^{(2n+1)}}{(2n +1)!}\bigg) \bigg(\sum_{n=0}^{\infty} (z^2)^n\bigg) \\ &= \bigg(\sum_{n=0}^{\infty} \frac{z^{2n - 3}}{(2n +1)!}\bigg) \bigg(\sum_{n=0}^{\infty} z^{2n-4} \bigg) \\\end{aligned} $$

I don't really know where to go from here. Any help would be great.


  • $\begingroup$ @oldrinb I don't know. If $|z| < 1$ I would effectively be saying that $|\sum_{n=0}^\infty z^{2n}| = \sum_{n=0}^\infty \frac{1}{z^{2n}} $ ? $\endgroup$ – franklin Apr 21 '13 at 5:37
  • $\begingroup$ Do you know how to multiply polynomials? Multiplying these power series is similar. $\endgroup$ – obataku Apr 21 '13 at 5:56
  • $\begingroup$ see my response... $\endgroup$ – obataku Apr 21 '13 at 8:06

For the product of the two series, we aim for matching powers of $z$ and then take the Cauchy product of our coefficients:

$$\begin{align*}\frac1{z^4}\left(\sum_{n=0}^\infty\frac{z^{2n+1}}{(2n+1)!}\right)\left(\sum_{n=0}^\infty z^{2n}\right)&=\frac1{z^3}\left(\sum_{n=0}^\infty\frac{z^{2n}}{(2n+1)!}\right)\left(\sum_{n=0}^\infty z^{2n}\right)\\&=\frac1{z^3}\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac1{(2k+1)!}\right)z^{2n}\\&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac1{(2k+1)!}\right)z^{2n-3}\end{align*}$$For our residue, we're interested in the residue for $z=0$, we want the coefficient of $z^{-1}=z^{2(1)-3}$, which corresponds to $n=1$:$$\operatorname{Res}_{z=0}\frac{\sinh z}{z^4(1-z^2)}=\sum_{k=0}^1\frac1{(2k+1)!}=\frac1{1!}+\frac1{3!}=\frac76$$

For $\frac{\cot z}{z^4}$, first we rewrite $\cot z=\frac{\cos z}{\sin z}$:$$\frac{\cot z}{z^4}=\frac{\cos z}{z^4\sin z}$$Recall our familiar series expansions:$$\cos z=\sum_{n=0}^\infty\frac{(-1)^n z^{2n}}{(2n)!}=1-\frac{z^2}2+\frac{z^4}{4!}-\frac{z^6}{6!}+\dots\dots\\\sin z=\sum_{n=0}^\infty\frac{(-1)^nz^{2n+1}}{(2n+1)!}=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\frac{z^7}{7!}+\dots$$Now, recognize that since we're interested in the residue for $z=0$, and our power series ends up divided by $\frac1{z^4}$, our coefficients are shifted backwards so we only need the coefficients for the first few terms of our power series. Let $\frac{\cos z}{\sin z}=a_{-1}z^{-1}+a_0z+a_1z^3+a_2z^5+a_3z^7+\dots$ so that:$$\left(z-\frac{z^3}{3!}+\frac{z^5}{5!}-\frac{z^7}{7!}+\dots\right)\left(a_{-1}z^{-1}+a_0z+a_1z^3+a_2z^5+a_3z^7+\dots\right)\\=1-\frac{z^2}2+\frac{z^4}{4!}-\frac{z^6}{6!}+\dots\\a_{-1}+a_0 z^2+a_1z^4+a_2z^6-\frac{a_{-1}}{3!}z^2-\frac{a_0}{3!}z^4-\frac{a_2}{3!}z^6+\frac{a_{-1}}{5!}z^4+\frac{a_0}{5!}z^6-\frac{a_{-1}}{7!}z^6+\dots\\=1-\frac{z^2}2+\frac{z^4}{4!}-\frac{z^6}{6!}+\dots\\a_{-1}+\left(a_0-\frac16a_{-1}\right)z^2+\left(a_1-\frac16a_0+\frac1{120}a_{-1}\right)z^4+\left(a_2-\frac16a_1+\frac1{120}a_0-\frac1{5040}a_{-1}\right)z^6+\dots\\=1-\frac{z^2}2+\frac{z^4}{4!}-\frac{z^6}{6!}+\dots$$ Equate coefficients for the first several terms:$$a_{-1}=1\\a_0-\frac16a_{-1}=-\frac12\implies a_0=-\frac13\\a_1-\frac16a_0+\frac1{120}a_{-1}=\frac1{24}\implies a_1=-\frac1{45}\\a_2-\frac16a_1+\frac1{120}a_0-\frac1{5040}a_{-1}=-\frac1{720}\implies a_2=-\frac2{945}$$Given this, we can write our power series for $\cot z$ as follows:$$\cot z=z^{-1}-\frac13z-\frac1{45}z^3-\frac2{945}z^5+\dots$$Now, divide through both sides by $z^4$ to yield our desired series:$$\frac{\cot z}{z^4}=z^{-5}-\frac13{z^{-3}}-\frac1{45}z^{-1}-\frac2{945}z+\dots$$Our residue is then merely the coefficient of our $z^{-1}$ term:$$\operatorname{Res}_{z=0}\frac{\cot z}{z^4}=-\frac1{45}$$


A related problem. Lets consider your first problem

$$ \frac{\cot(z)}{z^4}=\frac{\cos(z)}{z^4\sin(z)}. $$

First, determine the order of the pole of the function at the point $z=0$, which, in this case, is of order $5$. Once the order of the pole has been determined, we can use the formula

$$r = \frac{1}{4!} \lim_{z\to 0} \frac{d^4}{dz^4}\left( z^5\frac{\cos(z)}{z^4\sin(z)} \right)=-\frac{1}{45}. $$

Note that, the general formula for computing the residue of $f(z)$ at a point $z=z_0$ with a pole order $n$ is

$$r = \frac{1}{(n-1)!} \lim_{z\to z_0} \frac{d^{n-1}}{dz^{n-1}}\left( (z-z_0)^n f(z) \right) $$

Note: If $z=z_0$ is a pole of order one of $f(z)$, then the residue is

$$ r = \lim_{z\to z_0}(z-z_0)f(z). $$

  • 1
    $\begingroup$ I think he intends for it to be derived explicitly using a Laurent series, but nice work! $\endgroup$ – obataku Apr 21 '13 at 7:36
  • $\begingroup$ @oldrinb: Thanks for the comment. $\endgroup$ – Mhenni Benghorbal Apr 21 '13 at 12:50

$$\cos z=1-\frac{z^2}{2}+\frac{z^4}{24}-\ldots\;,\;\;\;\sin z=z-\frac{z^3}{6}+\frac{z^5}{120}-\ldots\implies$$

$$\cot z=\frac{\left(1-\frac{z^2}{2}+\frac{z^4}{24}-\ldots\right)}{z\left(1-\left(\frac{z^2}{6}-\frac{z^4}{120}\right)+\mathcal O(z^6)\right)}=$$



$$=\frac{1}{z}\left(1-\frac{z^2}{3}-\frac{z^4}{45}+\ldots\right)=\frac{1}{z}-\frac{z}{3}-\frac{z^3}{45}+\mathcal O(z^5)$$

Multipying the above by $\,\displaystyle{\frac{1}{z^4}}\;$ renders the residue $\,\displaystyle{-\frac{1}{45}}\,$ .

Notice we only use the powers necessary to calculate the coefficient of $\,z^{-1}\,$ . All the rest is unimportant to us.


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