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This is a question of an old exam of my complex analysis course and although I think I understand what a singularity is, I oftentimes have troubles 'finding and describing' them properly. I looked at a lot of singularity-problems at this great site, and I have the feeling that singularities can be found and described with various methods.. I usually don't know what method to use! And this one I'm totally stuck:

Find and describe the singularities of the function

$$f(z) = \frac{e^{1/z}\cdot \cos(\sqrt{1-z^2}) - 1}{\sin(2\pi z)}.$$

This is what I thought:

$\sin(2\pi z) = 0$ for $z$ an integer! (order $1$, because the derivative of the $\sin$ at $z \in \mathbb{Z}$ isn't $0$).

Now you have to look at what happens when you put $z = k \in \mathbb{Z}$ in the numerator. But now this is totally confusing, because the root of negative numbers doesn't exist (branch cut at the negative reals)! So what to do then?

And if you try to make a Laurent series I have some serious trouble because the series for the denominator is $2\pi z - ((2\pi z)^3)/3! + ((2\pi z)^5)/5! + ... + \ldots$

And for the numerator I don't know how to make the series (because of the chain of functions), and even if I would know them, I don't know how to divide a series by another series!

Is there anyone who could help me? Is one of these two directions the way to go forward with at all?

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  • $\begingroup$ Welcome to MSE! It really help readability to format using MathJax (see FAQ). It is always helpful to state where this problem is from and what you have tried. Regards $\endgroup$
    – Amzoti
    Jul 1, 2013 at 15:27
  • $\begingroup$ Do you know the types of singularities? $\endgroup$ Jul 1, 2013 at 16:16
  • $\begingroup$ Thanks @Amzoti ! And also thanks for reformatting Mhenni Benghorbal. Should I be more specific about where I got the problem from then now stated? And I stated what I've tried so far! (at least: 'thoughts') I know it isn't much but that's as far as I can get! $\endgroup$ Jul 1, 2013 at 16:19
  • $\begingroup$ I know the types @mhenni-Benghorbal, you have removable singularities, poles of finite order and essential singularities. I know $e^(1/z)$ has an essential singularity at z=0, and $sin(2*pi*z)$ has zeros of order 1 at z = 0. But since there are chained functions, a series expansion seems impossible to me, and I don't know how to cope with the branch cut of the root in the cosine! Or isn't that the way forward? $\endgroup$ Jul 1, 2013 at 17:01

1 Answer 1

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The singularities of $f$ are

  • the singularities of the numerator $e^{1/z}\cdot \cos \sqrt{1-z^2} - 1$, and
  • the zeros of the denominator $\sin (2\pi z)$.

In general, singularities of the denominator would have to be considered too, but here the denominator is entire.

The zeros of the denominator are exactly the numbers $z_k := \frac12\cdot k,\; k \in \mathbb{Z}$. All these are simple zeros, since the derivative, $2\pi\cos (2\pi z)$, is $\pm 2\pi$ at these points.

So $f$ has

  • a removable singularity at $z_k$ if the numerator has a zero at $z_k$,
  • a simple pole at $z_k$ if the numerator is holomorphic at $z_k$ and does not vanish there,
  • an essential singularity at $z_k$ if the numerator has an essential singularity at $z_k$.
  • a singularity of type $T$ at each point $w \notin \{z_k \colon k \in \mathbb{Z}\}$ where the numerator has a singularity of type $T$.

Let's look at the numerator.

First, take a look at $\cos \sqrt{1-z^2}$. Now, the square root by itself is problematic, but that problem is solved by applying $\cos$ to it. $\cos$ is an even function, and the possible values of the square root differ only by sign, so $\cos \sqrt{g(z)}$ is an entire function for any entire function $g$ (from the power series expansion, you get

$$\cos\sqrt{g(z)} = \sum_{\nu=0}^\infty \frac{(-1)^\nu}{(2\nu)!}\left(\sqrt{g(z)}\right)^{2\nu} = \sum_{\nu = 0}^\infty \frac{(-1)^\nu}{(2\nu)!}g(z)^\nu$$

and the sum is locally uniformly convergent, hence the limit is holomorphic).

So that one is an entire function.

$e^{1/z}$ has an essential singularity at $0$, and no other singularities. The only way to get rid of an essential singularity is by multiplying with the constant $0$, subtracting a function with the same principal part, or multiplying with/dividing by a function with a fitting essential singularity. Here no other essential singularities occur, hence $f$ has an essential singularity at $0$.

The numerator has no further singularities, hence the only question that remains is whether some of the zeros $z_k,\, k \neq 0$ of the denominator are cancelled by zeros of the numerator. Since $\lvert z_k\rvert \geqslant 1$ for $\lvert k\rvert \geqslant 2$, the $\cos$ factor takes the value $\cos (i\sqrt{z_k^2-1}) = \cosh \sqrt{z_k^2-1} \geqslant 1$ there. For $k > 0$, $e^{1/z_k} > 1$ and therefore the numerator is positive (where for $k = 1$, we compute directly, since that is not covered by the general case). For $k < -3$, it is an easy estimate to see that the numerator takes a positive value in $z_k$. For $k \in \{-1,-2,-3\}$, a computation shows that the numerator takes a negative value.

Thus $f$ has

  • an essential singularity in $0$, and
  • simple poles in $z_k$ for $k \neq 0$.
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  • $\begingroup$ Thank you very much, I think I understand it now. I would like to upvote your answer but I don't have 15 reputation yet! Just one question springs to mind now: If there is a singularity in the denominator, does it mean the function there has a zero? For instance $1/(1/z)$ ? $\endgroup$ Jul 2, 2013 at 8:34
  • $\begingroup$ If the singularity in the denominator is a pole, and the numerator is holomorphic there (or has a removable singularity), or has a pole of smaller order, then the function has a zero there (after removing the removable singularity). If the numerator gas a pole of equal order, the removable singularity that results leads to a non-zero value there, if the numerator has a pole of greater order, the function will have a pole. If the numerator has an essential singularity, the overall result will be an ess. sing. If the denominator has an essential singularity, the function will, unless they cancel $\endgroup$ Jul 2, 2013 at 9:46
  • $\begingroup$ Clear as water (that's probably not an English expression), I thank you very much! $\endgroup$ Jul 2, 2013 at 15:59

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