@Grognor So i^i = (e^(i pi/2))^i = e^(i pi/2 i) = e^(-pi/2). Clear as mud, but that's how you do it.
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@Meaningness@Grognor Or at least the 1982985235 identities that are minor variations on sin^2 + cos^2 = 1. - End of conversation
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. Banned in Sweden. SubGenius, Zhuangist, white-hat troll. Defrocked mathematician. Brain problems.