0

I have been trying to determine the singularities of $\frac{z}{\cos z}$:

I tried this problem by finding the series $\frac{1}{\cos (z)}$ separately, and then multiplying like: $$z*\sum_{n=0}^{\infty} a_n z^{2n}$$.

Since we are not concerned with the coefficients, $a_n$, this tells me that there is no singularity at all, because we only have positive powers of $z$. So what types of singularities are these/this?

Other resources indicate that my expansion of 1/ cos z is correct: Find series expansion of 1/cosx

But I'm also other answers different from mine:

1) In this textbook on page 204: An Introduction to Complex Analysis by Agarwal, it says that the answer is $\frac{\pi}{2} + \pi$

Which is correct? My answer? or $\frac{\pi}{2} + \pi$ in Agarwal?

mrf
  • 43,639
makansij
  • 1,583
  • 2
    You picked up the wrong series from the links. That's the series of $z\cos(z)$. Now, even when you get the series of $z/\cos(z)$ centred at $z=0$ that series is not going to converge on the whole plane. Therefore, it will only tell you that $z/\cos(z)$ is analytic in some open disc centred at $z=0$. – conditionalMethod Dec 06 '19 at 16:45
  • Wow you're right! Sorry about that! – makansij Dec 06 '19 at 16:49
  • thanks, but how do we see that from the Laurent series? – makansij Dec 06 '19 at 16:50
  • 1
    Laurent series can tell you the type of singularity (in particular the order in the case of poles) for the point at which the expansion is centred. You will would need to use many series expansions at each of the points where $\cos(z)$ vanishes. But note that to detect poles and their orders you don't really need the series expansion. You only need that $\lim_{z\to a}\frac{z}{\cos(z)}=\infty$ and there is an integer $m>0$ such that $\lim_{z\to a}(z-a)^m\frac{z}{\cos(z)}=0$. – conditionalMethod Dec 06 '19 at 16:52
  • 1
    No, your expansion is for $\cos z,$ not $1/\cos z.$ – zhw. Dec 06 '19 at 17:50
  • @conditionalMethod, I understand you can compute the residue of the pole (I assume that is what you're getting at). But:
    1. This is painstaking because you don't know the order ahead of time
    2. This only works for poles of finite order (i.e. it will not work if there is an essential singularity).
     – makansij Dec 06 '19 at 23:55
  • Also, @conditionalMethod, I tried your method with $m=1$, and I got the limit to be $0$, as you said. But notice that if I had tried $m=2$, I also would have gotten zero. So, how would this method help me determine the order of the pole if I would get different values of $m$ that could be valid? – makansij Dec 07 '19 at 00:52
  • 1
    The order is one less than the smallest $m$ that makes the limit $0$. – conditionalMethod Dec 07 '19 at 01:13
  • But if $m=1$ is giving me a result of $0$, then how does that help me? – makansij Dec 07 '19 at 20:43
  • @conditionalMethod, to respond to your first comment, my understanding was that $\frac{1}{\cos z}$ converges for all $z$, so how do I know about which point to center the series. Is that true, or no? – makansij Dec 07 '19 at 21:49
  • 1
    Quotients of analytic functions, where the denominator is non-zero, are analytic. Therefore, the only points that need to be studied are those where $\cos(z)=0$. Actually all of them can be studied at once. – conditionalMethod Dec 07 '19 at 23:26
  • Thanks - believe it or not you are helping me tremendously. Almost there: If I just set the denominator of $f(z)$ in the question equal to 0, and solve for values of $z$, then yes I get the singularities in this case. However that doesn' always work - right? WHen do I need to expand into Laurent series and when can I just set denominator equal to $0$ and solve? (and furthermore, if I do expand in Laurent series how do I know around which point?) – makansij Dec 07 '19 at 23:45

1 Answers1

2

Hint: Do the following: 1. Find the zeros of $\cos z.$ 2. Show that these are simple zeros of $\cos z.$ 3. Show that if $f,g$ are holomorphic near $a,$ $g$ has a simple zero at $a,$ and $f(a)\ne 0,$ then $f/g$ has a simple pole at $a.$

zhw.
  • 105,693