Is z=0 a Non-Isolated Singularity in \(\frac{1}{\sin(\frac{\pi}{z})}\)?

[krindik]

Hi,
The function $\frac{1}{\sin(\frac{\pi}{z})}$ has isolated singularities at z=+-1, +-1/2, ...

However, it is said that it has an non-isolated singularity at z=0.
A non-isolated singularity has to be a point where its neigborhood too is also singular.

But, for some $\epsilon > 0 ,\, \frac{1}{\sin(\frac{\pi}{\epsilon})}$ is not singular eg. $\epsilon = 0.000001$

Can u pls explain.

Thanks

 

[elect_eng]

Hi,
The function $\frac{1}{\sin(\frac{\pi}{z})}$ has isolated singularities at z=+-1, +-1/2, ...

However, it is said that it has an non-isolated singularity at z=0.
A non-isolated singularity has to be a point where its neigborhood too is also singular.

But, for some $\epsilon > 0 ,\, \frac{1}{\sin(\frac{\pi}{\epsilon})}$ is not singular eg. $\epsilon = 0.000001$

Can u pls explain.

Thanks

My guess is that this function has a branch cut and in the neighborhood of z=0, there are singularities at the branch cut, as well as exactly at zero.

Note that I haven't studied that function carefully, so it's just a guess.

I've had trouble with this concept too. It's really confusing.

Here is a link to an explanation.

http://en.wikipedia.org/wiki/Branch_cut

I've also attached a PDF which talks about this a little.

 

[krindik]

Thanks.
Isn't branch cuts used for multi-valued functions? ie. to map multi-valued functions to single-valued functions on Reimann sheets?

 

[elect_eng]

Thanks.
Isn't branch cuts used for multi-valued functions? ie. to map multi-valued functions to single-valued functions on Reimann sheets?

Yes. A simple example is natural logarithm log(z). The negative real axis is used as an arbitrary cut to define a single valued function. This then creates a singularity so that no matter how small the neighborhood around z=0, there is the singularity at zero and at the branch cut a small distance away from z=0. I'm guessing your function has the same issue, but I have not verified that.

 

[CRGreathouse]

Hi,
The function $\frac{1}{\sin(\frac{\pi}{z})}$ has isolated singularities at z=+-1, +-1/2, ...

However, it is said that it has an non-isolated singularity at z=0.
A non-isolated singularity has to be a point where its neigborhood too is also singular.

But, for some $\epsilon > 0 ,\, \frac{1}{\sin(\frac{\pi}{\epsilon})}$ is not singular eg. $\epsilon = 0.000001$

Yes, there's a neighborhood around 0.000001 where everything's fine. But within a circle of radius 0.000001 around zero there are infinitely many singularities.

 

[krindik]

Thanks,

Yes, there's a neighborhood around 0.000001 where everything's fine. But within a circle of radius 0.000001 around zero there are infinitely many singularities.

But how can there be a singularity in $0 < |z| < 0.000001$ ? eg. at $\frac{0.000001}{2}$ because $\frac{1}{\sin(\frac{\pi}{\frac{0.000001}{2}})}$ is not singular?

Can u pls elaborate?

 

[elect_eng]

The branch-cut concept I mentioned may apply here, but I see CRGreathouse's point. The mapping of the real number line, via pi/z, into the sine function creates an infinity of zeros inside any finite radius that is established. This is a much simpler way to show that the point is a non-isolated singularity.

 

[squidsoft]

It's a non-issolated singularity and no branch cuts are involved: $$\csc(a/z)$$ is single-valued. The origin is a non-issolated singularity because there is no deleted neighborhood of the origin, $$D(0,\delta)$$, however small, that contains only one singularity. I think also it's a bonifide non-issolated, essential singularity, that is, in any deleted neighborhood of the origin, the function assumes every value in C infinitely often.

 

[elect_eng]

It's a non-issolated singularity and no branch cuts are involved: $$\csc(a/z)$$ is single-valued.

This function seems to have the same issue as the log(z) function in that angles with multiples of 2pi do not give the same answer. See the following calculations from Matlab. Since log(z) is considered to have a branch cut on the negative real axis, to have single-values, should not this function also?

>> 1/sin(-pi/.0001)

ans =

2.4250e+011

>> 1/sin(pi/.0001/exp(-i*pi))

ans =

1.2965e+011 -1.2096e+011i

>> 1/sin(pi/.0001/exp(i*pi))

ans =

1.2965e+011 +1.2096e+011i

The fact that there are three different answers above shows that the function is not single valued unless a branch cut is used to define a modified function that is single valued, this then generates a discontinuity at the branch cut.

 

[krindik]

Thanks.

I got ur point. What u r saying is the given function can get any value within the neighborhood of z=0.

This can be easily visualized for example by $e^{\frac{1}{z}}$
Representing z in polar form we get,
$e^{\frac{1}{z}} = A e^{\imath\alpha}$

We can represent $r, \theta$ in terms of $A, \alpha$ and by keeping $A = const, \alpha=2\pi n, n=\pm{1}, \pm{2}, ...$ we can get many values for r

Note also the Picard's theorem : if an analytic function f(z) has an essential singularity at a point w then on any open set containing w, f(z) takes on all possible complex values, with at most a single exception, infinitely often.