A point $z_0$ is called a singularity of a complex function $f$ if $f$ is not analytic at $z_0$, but every neighborhood of $z_0$ contains at least one point at which $f$ is analytic.
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A singularity $z_0$ of a complex function $f$ is said to be a point singularity or isolated singularity if there exists a neighborhood of $z_0$ in which $z_0$ is the only singularity of $f$.
Alternatively, a point singularity of a function $f$ is a @complex-number $z_0$ such that $f$ is defined in a neighborhood of $z_0$ but not at the point $z_0$ itself.
Classification of Singularities
Given a @Laurent-series expansion of a function with an isolated singularity at $z_0$, we have the possibility of the singularity being removable, a pole, or an essential singularity.
A removable singularity occurs if all negative powers in the Laurent series are zero and the function can be redefined at the singularity to be analytic.
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Consider $f(z) = z$ defined in the @punctured-plane. Then, the @origin is an isolated singularity, because $f(z)$ is not defined there, but is defined everywhere else in a neighborhood of the origin. We can easily define $f(0) = 0,$ and this will make $f(z)$ analytic at $0,$ and so we say this singularity is removable.
A pole is present if the Laurent series has a finite number of negative power terms. The largest negative exponent (in absolute value) indicates the order of the pole.
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Consider $f(z) = 1/z$ defined in the @punctured-plane. Then, the @origin is an isolated singularity, but in this case, we can't simply define it in a continuous fashion, because $f(z)$ approaches infinity as $z \to 0.$ Thus, we call this singularity a pole.
An essential singularity occurs if there are infinitely many negative powers in the Laurent series.
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Consider $f(z) = e^{1/z},$ on the @punctured-plane. As $z$ approach $0$ on the positive real axis, $f(z)$ goes to infinity, and as $z$ approaches $0$ on the negative real axis, $f(z)$ approaches 0. On the imaginary axis, $f(z)$ oscillates wildly, but remains bounded, as $z$ approaches $0.$ This is neither a removable singularity nor a pole, and we call it an essential singularity.
If $f(1/z)$ has a singularity at $0,$ then $f(z)$ has a singularity at the @point-at-infinity.