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For the function \(f\left( z \right) = \sin \left( {\frac{1}{z}} \right),\;z = 0\) is a:
1. Removable singularity
2. Simple pole
3. Branch point
4. Essential singularity

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Correct Answer - Option 4 : Essential singularity

\(f\left( z \right) = \sin \left( {\frac{1}{z}} \right)\)

We know that,

\(\sin z = z - \frac{{{z^3}}}{{3!}} + \frac{{{z^5}}}{{5!}} \ldots\)

\(f\left( z \right) = \sin \left( {\frac{1}{z}} \right) = \frac{1}{z} - \frac{1}{{3!{z^3}}} + \frac{1}{{5!{z^5}}} \ldots\)

If the number of negative powers of (z - a) in the f(z) is negative, then z = a is called an essential singularity.

∴ f(z) has infinite negative powers of z = 0 and z = 0 is an essential singularity.

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