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jack action's latest activity
jack action
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TIL about Ewin Howard Armstrong:
Sep 2, 2025
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I learned that $$|e^{i\pi}|=|\pi^{ie}|=|i^{e\pi}|=1$$
Sep 2, 2025
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I learned that $$|e^{i\pi}|=|\pi^{ie}|=|i^{e\pi}|=1$$
