0
$\begingroup$
  1. What was the singularity type of $\frac{z\sin(z)}{\cos(z)-1}$?

  2. and what was its Laurent series expansion?

It was essential singularity. But I used $\displaystyle\frac{\sum_{n=0}^\infty z^{2n+1}/(2n+1)!}{\sum_{n=0}^\infty z^{2n+1}/(2n+2)!}\approx\frac{\alpha z+\gamma_1(z)}{\beta z+\gamma_2(z)}$ where it seemed that the limit of $f(z)$ exists, which was wrong. how to prove it, and what was its laurent series expansionn?

$\endgroup$
  • 1
    $\begingroup$ $\cos(z)-1$ vanishes at $z=0,\pm 2\pi,\pm 4\pi,\cdots$ and its derivative $\sin(z)$ also vanishes at these points. The zeros of $\cos(z)-1$ are of order 2. So $0$ is a removable singularity of $f$. All other zeros of $\cos(z)-1$ result in first order poles of $f$. $\endgroup$ – DisintegratingByParts Nov 14 '17 at 18:17
  • $\begingroup$ @DisintegratingByParts but how to prove it formally? Both the numerator and denominator gots to 0 at singularities... $\endgroup$ – user416486 Nov 14 '17 at 20:58
  • $\begingroup$ Expand $\cos(z)-1$ in a power series about $z=\pm 2n\pi$, and note that the function and its first derivative vanish at these points, but the second derivative does not. $\endgroup$ – DisintegratingByParts Nov 14 '17 at 23:10
0
$\begingroup$

For $z=0$ we have

$$ \begin{align} \lim_{z\to 0} \frac{z\sin z}{\cos z - 1} &= \lim_{z\to 0} \frac{z\left(z - \frac{z^3}{3!}+\cdots\right)}{\left(1 - \frac{z^2}{2} + \frac{z^4}{4!} - \cdots \right) -1} \\ &= \lim_{z\to 0} \frac{z^2 - \frac{z^4}{3!}+\cdots}{-\frac{z^2}{2} + \frac{z^4}{4!} - \cdots} \\ &= -2 \end{align} $$

making it a removable singularity


For $z = 2n\pi$

$$ \begin{align} \lim_{z\to 2n\pi} (z-2n\pi)\frac{z\sin z}{\cos z - 1} &= \lim_{w\to 0} \frac{w(w+2n\pi)\sin w}{\cos w - 1} \\ &= \lim_{w\to 0} \frac{w^2\sin w}{\cos w-1} + 2n\pi\lim_{w\to 0} \frac{w\sin w}{\cos w - 1} \\ &= 0 + 2n\pi(-2) \\ &= -4n\pi \end{align} $$

making it a pole of order 1 for $n\ne 0$


Another way to see this is $$ \frac{z\sin z}{\cos z - 1} = -\frac{2z \sin \frac{z}{2} \cos \frac{z}{2}}{2\sin^2 \frac{z}{2}} = -z\cot \frac{z}{2} $$

which has simple poles at $\frac{z}{2} = n\pi$ except the origin


For the Laurent series (I'm assuming for $|z|<2\pi$), you will need to know the series for $\cot z$. Depending on the problem requirements, you can solve it term-by-term or look up a closed-form.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you so much! This is beautiful. $\endgroup$ – user416486 Nov 16 '17 at 1:12

Your Answer

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