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Thus
Thus i^{i} is a real number! In decimal form it is approximately 0.207880. Here is a more algebraic way to see it. De Moivre showed that e^{ix} = cosx + isinx so that e^{iπ/2} = cos(π/2) + isin(π/2) = i, for example. Thus to look at i^{i} one could consider i^{i} = (e^{iπ/2})^{i} and that's the same as e^{(iπ/2)i} = e^{(i2)(π/2)} = e^{π/2}. Penny
Hi, Since sine and cosine are periodic with period 2π, for any integer n
Hence i^{i} has infinitely many values, e^{2nπ π/2} for n any integer. The valuse closest unity being approximately
RD  


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