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I have some quetion on essential singularity.

Let $f(z)$, $g(z)$ have the same essential singularity at $z=z_0$.

Then, if $\frac{f(z)}{g(z)}$ is not a constant function on some neighborhood of $z_0$, then $ \frac{f(z)}{g(z)}$ also has essential singularity at $z=z_0$?

If not, could you give me some counter example?

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    $\begingroup$ What does it mean for essential singularities to be the same? I've never encountered this idea. $\endgroup$
    – Cameron Williams
    Aug 30, 2013 at 5:45
  • $\begingroup$ @CameronWilliams I don't think that it is a real concept. If $f$ has an essential singularity at $z=0$, then $z\cdot f(z)$ does have the same (?) essential singulary but $\frac{z\cdot f(z)}{f(z)}=z$ $\endgroup$
    – Dominic Michaelis
    Aug 30, 2013 at 6:15
  • $\begingroup$ I think the OP just meant "also has essential singularity, at the same point" maybe? (in which case Dominic has a counterexample) $\endgroup$
    – Evan
    Aug 30, 2013 at 6:29
  • $\begingroup$ Thanks Dominic for your counterexample. It was so near to me. $\endgroup$
    – Andrew
    Aug 30, 2013 at 7:35

2 Answers 2

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Dominic Michaelis provided the following counterexample in the comments section: if $f$ has an essential singularity at $z=z_0$, then also $g(z)=zf(z)$ has an essential singularity at $z=z_0$ but $\frac{g(z)}{f(z)}=\frac{zf(z)}{f(z)}=z$ which does not have an essential singularity at $z=z_0$.

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I'm going to assume that by essential singularity you mean is 0 at the given point and that the singularity occurs when the function is in the denominator. The classic example for this is the sinc function, sin(x)/x. This is the Fourier transform for a rectangular window. Clearly both functions approach 0 at x = 0, but if you take the derivative of each you will see that x approaches 0 at a constant rate and sin(x) approaches 0 as a function of cos(x). The limit as x ->0 is 1, so that the value is non-singular and the function itself is analytic at all points of the curve.

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  • $\begingroup$ $\sin(x)/x$ has a removable singularity at $0$, not an essential one. $\endgroup$
    – mrf
    Aug 31, 2013 at 22:13
  • $\begingroup$ You're right. I was a bit sloppy in my reading. I was thinking of sinx and x separately, but even then they have a pole and zero and not an essential singularity. $\endgroup$
    – Dennis Fleming
    Sep 2, 2013 at 18:25

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