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  "body": "\n## Proof\n\nWe split $z$ into real and imaginary parts and use the periods of sine and cosine.\n\n1. Write $z = x + iy$ with $x, y$ real. By the [[product of exponentials|0WM7DJCM6G]], $e^z = e^x e^{iy}$, where $e^w$ is the [[complex exponential|054VXHMCA3]].\n2. At the real number $x$, the exponential series sums to the [[real exponential|06JZGMYTYD]]. So $e^x$ is a [[positive real number|0P9BNBW2W3]].\n3. By [[Euler's formula|04R3CY41VJ]], $e^{iy} = \\cos y + i \\sin y$. At real $y$, these are the real [[cosine|08W5RN375X]] and [[sine|02KTRHQMHN]]. So $\\cos y$ and $\\sin y$ are the real and imaginary parts of $e^{iy}$.\n4. By the [[definition of the modulus|0KW2X0TFBX]] and the [[Pythagorean identity|03H7WZ4T3K]], $\\abs{e^{iy}} = \\sqrt{\\cos^2 y + \\sin^2 y} = 1$.\n5. Let $k$ be an [[integer|0C4JS9RF5M]]. Applying the [[periodicity of sine and cosine|0P4NQD746K]] $\\abs{k}$ times gives $\\cos(2\\pi k) = \\cos 0 = 1$ and $\\sin(2 \\pi k) = \\sin 0 = 0$.\n6. So if $z = 2 \\pi i k$, then $e^z = e^{i (2\\pi k)} = \\cos(2\\pi k) + i \\sin(2\\pi k) = 1$ by step 3. This proves one direction.\n7. Conversely, assume $e^z = 1$. The [[modulus is multiplicative|0HYNZE940T]], so $1 = \\abs{e^z} = \\abs{e^x} \\abs{e^{iy}} = e^x$ by steps 2 and 4.\n8. The real exponential is [[strictly increasing|0KN2V7840E]] and $e^0 = 1$. So $e^x = 1$ forces $x = 0$.\n9. Then $1 = e^{iy} = \\cos y + i \\sin y$. Comparing real and imaginary parts, $\\cos y = 1$ and $\\sin y = 0$.\n10. Let $k = \\floor{y / (2\\pi)}$, an integer, and $t = y - 2\\pi k$. Then $0 \\le t \u003c 2\\pi$.\n11. By periodicity, as in step 5, $\\cos t = \\cos y = 1$ and $\\sin t = \\sin y = 0$.\n12. Suppose that $t \u003e 0$. For each real $s$, the [[angle addition formula|028XEQCE88]] gives $\\sin(s + t) = \\sin s \\cos t + \\cos s \\sin t = \\sin s$.\n13. So $t$ is a positive period of $\\sin$ with $t \u003c 2\\pi$. This contradicts the fact that $2\\pi$ is the smallest positive period of $\\sin$. Hence $t = 0$.\n14. So $y = 2 \\pi k$ and $z = x + iy = 2 \\pi i k$ with $k$ an integer.\n\n## Idea\n\nThe modulus of $e^{x + iy}$ is $e^x$ and its angle is $y$. So $e^z = 1$ forces the modulus $e^x$ to be $1$ and the angle $y$ to be a full number of turns. The minimality of the period $2\\pi$ rules out any other angle.\n",
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