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An isolated singularity of a complex function at $a$ is a missing or nonholomorphic point surrounded by a punctured disk $0<|z-a|<r$ on which the function is holomorphic. The punctured-disk hypothesis is decisive: it gives a Laurent series centered at $a$ and the three-way classification of isolated singularities.

At zero, $\sin z/z$ has a removable singularity, $z^{-2}$ has a double pole, and $e^{1/z}$ has an essential singularity. Their principal parts contain respectively no negative powers, finitely many negative powers, and infinitely many negative powers. For the first example, assigning the limiting value one fills the puncture holomorphically. For the second, multiplying by $z^2$ leaves a nonzero regular value. The third has the convergent punctured-plane expansion ${\sum }_{n\ge 0}{z}^{-n}/n!$.

An accumulation of other singularities changes the question. In reciprocal cosine with accumulating poles, every disk about zero contains poles, so zero is a non-isolated singularity and the same Laurent classification does not apply there. At infinity use the reciprocal coordinate $w=1/z$, as in pole at infinity and removable singularity at infinity.^[ref]

Use the local coordinate $w=1/z$ at infinity. For positive integer $N$,The negative powers of this local coordinate are precisely $m=0,\dots ,N-1$. The singular part at infinity is thereforeInfinity is a pole at infinity of order $N$. In the original $z$ variable these singular terms are positive powers; it is the negative powers of $w=1/z$ that define the local principal part of a Laurent series. For $N=0$ the point at infinity is regular, and for negative integer $N$ it is a zero rather than a pole.

The Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$. A holomorphic function is smooth, so equality of mixed partial derivatives givesThus its real and imaginary parts are harmonic functions.

An entire function is holomorphic at every finite point of the complex plane. The globally convergent power series for the exponential implies that $e^z$ and $e^{-z}$ are entire. Their difference therefore givesat every complex $z$, proving that the hyperbolic sine is entire. For powers, the base case $f_1(z)=z$ has derivative one everywhere. If $z^n$ is entire with derivative $n{z}^{n-1}$, the product rule gives $(z^{n+1}{)}^{\prime }=z^n+zn{z}^{n-1}=(n+1)z^n$ everywhere. Induction proves that every positive integer power is entire.

A function has a pole of order $n$ at $z_0$ when it is holomorphic in a punctured neighbourhood and $g(z)=(z-z_0{)}^{n}f(z)$ extends holomorphically to $z_0$ with $g(z_0)\ne 0$. An isolated essential singularity is neither a removable singularity nor a pole; equivalently its Laurent series has infinitely many nonzero negative-power terms. In particularhas an essential singularity at zero.

For $f_N$, the Laurent series about infinity isTo classify the point correctly, use $w=1/z$. Then $f_N(1/w)=w^{-N}{e}^w$, and $e^w$ is holomorphic with value one at zero. ThereforeThe pole at infinity is compatible with an essential singularity at the finite point zero. Negative powers of $z$ are regular powers of the local coordinate $w$ near infinity, so they do not make that point essential.