{
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  "body": "\n## Statement\n\nFor each $z \\in \\mathbb{C}$,\n$$e^{z + 2 \\pi i} = e^{z},$$\nwhere $e^z$ is the [[complex exponential|054VXHMCA3]] of $z$.\n\n## Remarks\n\nProof: by [[Euler's formula|04R3CY41VJ]], $e^{2 \\pi i} = \\cos 2\\pi + i \\sin 2\\pi = 1$. By the [[product of exponentials|0WM7DJCM6G]], $e^{z + 2\\pi i} = e^z e^{2\\pi i} = e^z$.\nThe periods of $\\exp$ are exactly the numbers $2 \\pi i k$ with $k \\in \\mathbb{Z}$, by [[exp z = 1 iff z is in 2πiZ|013AHC4DRT]].\nSo $\\exp$ is not injective on $\\mathbb{C}$. It is injective on each horizontal strip $\\{z : a \u003c \\operatorname{Im} z \\le a + 2\\pi\\}$ with $a \\in \\mathbb{R}$.\n\n## Explanations\n\n### L1 human\n\nAdding $2 \\pi i$ to $z$ adds a full turn to the direction of $e^z$ and does not change its size.\nSo $e^z$ takes the same value again: the exponential repeats itself along each vertical line, with step $2\\pi$.\n",
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  "name": "The complex exponential is 2πi-periodic",
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