{
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  "body": "\n## Example\n\nLet $f(z) = e^{1/z}$ for $z \\in \\mathbb{C} \\setminus \\{0\\}$, where $e^w$ is the [[complex exponential|054VXHMCA3]].\nThe function $f$ is holomorphic on $\\mathbb{C} \\setminus \\{0\\}$, so $0$ is an isolated singularity of $f$.\nIts [[Laurent series|0X3XD2ATB0]] at $0$ is\n$$e^{1/z} = \\sum_{n=0}^{\\infty} \\frac{1}{n!} z^{-n} = 1 + \\frac{1}{z} + \\frac{1}{2! \\, z^2} + \\cdots \\qquad (z \\neq 0).$$\nIt has infinitely many nonzero terms with negative powers, so $0$ is an [[essential singularity|0YKDJBV26S]] of $f$.\n\n## Remarks\n\nA direct check: for real $x \\to 0^+$, $e^{1/x} \\to \\infty$, and for real $x \\to 0^-$, $e^{1/x} \\to 0$. So $f$ has no limit at $0$ and $|f|$ does not tend to $\\infty$.\nIn each punctured disc $0 \u003c |z| \u003c \\delta$, $f$ takes each value $w \\neq 0$ infinitely often: the solutions of $e^{1/z} = w$ are $z = 1 / (\\operatorname{Log} w + 2\\pi i k)$ with $k \\in \\mathbb{Z}$ and $\\operatorname{Log} w + 2\\pi i k \\neq 0$, and these points tend to $0$ as $|k| \\to \\infty$.\nThe value $0$ is never taken. This agrees with [[Picard's great theorem|08QA8QTM71]], which allows one exception.\nThe residue of $f$ at $0$ is the coefficient of $z^{-1}$, which is $1$.\n\n## Explanations\n\n### L1 human\n\nNear $0$, the number $1/z$ is huge and can point in any direction, so $e^{1/z}$ can be huge, tiny, or anything in between.\nThat is why no single value or infinity describes $e^{1/z}$ near $0$.\n",
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