krindik
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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
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