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Here are a few functions for reference purposes:

$f(z) = \frac{sin(2z)}{z^3}$, $ \space g(z) = \frac{sin(z)}{tan(z)}$, $ \space h(z) = z^2 sin(\frac{1}{z})$

Suppose I was calculating the singularities of these functions at $z_0 = 0$ and I had to find the order of each function at $z_0$

I understand what all the different singularities are:- removable, pole and essential.

If I plot the graph of say $f(z)$ in the real plane where $z$ is a real number, it looks like the singularity is a pole. This is only a visual guess. It's definitely not removable.

It seems correctable ("fixable"), by multiplying $f$ by $\frac{z^3}{sin(2z)}$, but this would imply any bijective function with a singularity that is not removable must be a pole, right?

I don't think the function is essential, but my understanding of essential singularities is very weak.

When it comes to orders of a function at $z_0$, I know how to factorize polynomials and determine the order but I'm a bit stuck on the trigonometric functions.

Do I need to represent them as a polynomial using Taylor Series, or Laurent Series?

Also, is it a bad idea visualizing these complex functions as real functions.

If anyone has any better examples for explanation purposes, that's fine, I only have the functions above for referencing in the question.

Thanks in advance for any help.


My guesses for $g$ and $h$ are pole and essential respectively.

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  • $\begingroup$ By representing the functions as their unique Laurent series, you can find the order easily (the order is the infimum $n \in \mathbb{Z}$ of the non-zero $c_n$'s of the Laurent series). Also, if you have the order, you can figure out the type of the singularity. $\endgroup$ May 1, 2016 at 19:34

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A singularity of a holomorphic function $w$ at a point $a$ is:

  • Removable if the limit of $w$ at $a$ exists, and if $w$ were extended to have the limit as its value at $a$, then the extended function is holomorphic at $a$. ("Holomorphic at $a$" = "analytic at $a$" = "derivative exists in some neighborhood of $a$".)
  • A pole, if it is not removable, but there is a positive integer $n$ such that $(z - a)^nw(z)$ has a removable singularity at $a$. The order of the pole (not the function) is the smallest value of $n$ that works.
  • Essential if it is not removable nor a pole.

I.e.,

  • A removable singularity is one that doesn't really need to be there. It is just an artifact of how the function is defined, not a natural part of the function's behavior.
  • A pole is a real singularity, but one that is well-behaved. In some neighborhood of the singularity, the function acts like $\frac1{z^n}$ does near $0$.
  • An essential singularity is not well-behaved. Unlike poles, it may not be isolated, and even when it is isolated, its behavior is complicated (per Picard's theorem, it is very complicated).

For your functions,

  • Note that multiplying $f(z)$ by $z^3$ gives a function with an obviously removable singularity, so $f(z)$ has a pole. The order of that pole is at most $3$, but in fact it is smaller (I'll let you try to figure out what it really is).
  • For $g(z)$, recall that $\tan z = \frac {\sin z}{\cos z}$ and simplify. How does $\cos z$ behave at $0$? (Note that $g(z)$ itself is not defined at $0$, so it does have a singularity there.)
  • While $\lim_{z\to0} h(z)$ exists, consider $\lim_{z\to0} h'(z)$. If it is analytic, all derivatives of $h$ would exist. Multiplying $h(z)$ by $z^n$ just puts off the derivative where the limit fails to exist. It does not solve the issue altogether.

Your concept on $f$ that it is fixable by multiplying by $\frac {z^3}{\sin 2z}$ is of no value. Every singularity of any analytic function is "fixable" by multiplying by the function's inverse! This is not a useful property.

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  • $\begingroup$ Thanks for the answer. I've a few questions. Isn't $g(z)$ basically $cos(z)$, in which case it has removable singularities at $\frac{\pi}{2}+n\pi$ ? But none at $0$? And for $h$, if the limit exists then it's not essential, so it has to be a pole or removable, right? But which one, and how do we know? $\endgroup$ May 2, 2016 at 11:52
  • $\begingroup$ $g(z)$ is basically $\cos z$, but the "basically" is glossing over that they are defined on different sets. $\cos z$ is entire - it has no singularities. Equalling $0$ is not a singularity. $g(z)$ is not defined both when $\tan z$ is not defined, and when $\tan z = 0$, which includes $0$ itself. As for $h$, the limit existing is not sufficient to show that it is analytic at that point. To be removable, it has to be definable in such a way that it is analytic there - not possible. If it were a pole, then the limit would not exist here - not a pole either. So it is essential. $\endgroup$ May 2, 2016 at 14:57
  • $\begingroup$ Thanks. I didn't realize for it to be a pole, the limit could not exist $\endgroup$ May 2, 2016 at 15:14
  • $\begingroup$ When it is a pole, the limit is $\infty$ (i.e., the modulus (absolute value) of the function increases without bound). If we are considering $\Bbb C$ as the codomain of the function, then it doesn't exist. (If we consider the Reimann sphere as the codomain, then it does exist, as the Riemann sphere includes $\infty$ as a point.) $\endgroup$ May 2, 2016 at 15:54

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