Is this an example of an essential singularity?

**davismj** · March 3rd 2010, 07:34 PM

$f(z) = \frac{z^2}{sin z}$ with $z_0 = n\pi$ for $n = 0, \pm 1, \pm 2, . . .$

My reasoning is that clearly the function is undefined at the above defined singularities. Some of these are bounded, some of them are not.

**CaptainBlack** · March 3rd 2010, 11:12 PM

Originally Posted by davismj

$f(z) = \frac{z^2}{sin z}$ with $z_0 = n\pi$ for $n = 0, \pm 1, \pm 2, . . .$

My reasoning is that clearly the function is undefined at the above defined singularities. Some of these are bounded, some of them are not.

$z=0$ is a removable singularity, and all the others are poles.

CB

**davismj** · March 3rd 2010, 11:31 PM

Originally Posted by CaptainBlack

$z=0$ is a removable singularity, and all the others are poles.

CB

Actually, I think I might disagree with you. Here's why:

If $\displaystyle \lim_{z\to z_0}|f(x)|$ is bounded, the singularity is removable.

Therefore, by L'Hopital's rule we have.

$\displaystyle\lim_{z\to z_0}\frac{z^2}{sin z} = \displaystyle\lim_{z\to z_0}\frac{2z}{cos z} = 2z_0$

Which implies that the singularities are all removable, and their values are $2\pi n$ where $n = 0,\pm 1,\pm 2, . . .$

Does that sound right?

**CaptainBlack** · March 4th 2010, 12:02 AM

Originally Posted by davismj

Actually, I think I might disagree with you. Here's why:

If $\displaystyle \lim_{z\to z_0}|f(x)|$ is bounded, the singularity is removable.

Therefore, by L'Hopital's rule we have.

$\displaystyle\lim_{z\to z_0}\frac{z^2}{sin z} = \displaystyle\lim_{z\to z_0}\frac{2z}{cos z} = 2z_0$

Which implies that the singularities are all removable, and their values are $2\pi n$ where $n = 0,\pm 1,\pm 2, . . .$

Does that sound right?

No it does not, it only works for the singularity at the origin.

If you put $u=z-n\pi$ the function near the $z-n\pi$ singularity behaves as:

$\pm \frac{u+n\pi}{\sin(u)}$

(the sign depends on the parity of $n$ ) and the singularity behaves like $1/u$

CB

**shawsend** · March 4th 2010, 06:11 AM

Originally Posted by davismj

$f(z) = \frac{z^2}{sin z}$ with $z_0 = n\pi$ for $n = 0, \pm 1, \pm 2, . . .$

My reasoning is that clearly the function is undefined at the above defined singularities. Some of these are bounded, some of them are not.

Look for the spider (Yeah, well he's sideways in the plot below) or one of his close cousins. Here's the Mathematica code to draw the essential singularity of $\sin(1/z)$ . Try and understand why this code does what it does. Now substitute your function: no spider or even one of his relatives so no essential singularity.

Code:

spiderSingularity = ContourPlot[
   Re[Sin[1/(x + I*y)]] == 5, 
   {x, -0.1, 0.1}, {y, -0.1, 0.1}, 
   ContourStyle -> Red, PlotPoints -> 50]

**davismj** · March 4th 2010, 07:39 AM

Originally Posted by CaptainBlack

No it does not, it only works for the singularity at the origin.

If you put $u=z-n\pi$ the function near the $z-n\pi$ singularity behaves as:

$\pm \frac{u+n\pi}{\sin(u)}$

(the sign depends on the parity of $n$ ) and the singularity behaves like $1/u$

CB

I don't really understand your reasoning, but I guess it does make sense that the limit as say z approaches 1 is not $2\pi$ since z^2 approaches 1 and 1/sin z approaches infinity.

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