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Goal: Compute the possible values of $i^i$.

Attempt:

  1. We have that

    $$ i^i = e^{i \log(i)} $$

    by definition.

  2. Yet $\log(i)$ has an infinite number of values (if we aren't looking at its principal branch). That is,

    $$ \log(i) = k 2\pi i + {\pi \over 2}i \text{ for all }k \in \mathbb{Z} $$

Thus $i^i = {\pi i \over 2}$ in its principal branch, while all of its branches can be expressed as $k 2 \pi i + {\pi i \over 2}$ for all values of $k$.

Is this correct?

user1770201
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  • Other duplicates: https://math.stackexchange.com/questions/9776/how-to-raise-a-complex-number-to-the-power-of-another-complex-number https://math.stackexchange.com/questions/9770/understanding-imaginary-exponents https://math.stackexchange.com/questions/39913/complex-exponents – MJD Apr 29 '14 at 15:32

1 Answers1

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Not quite. You've found the values of $\log(i)$ - but you never went back and computed $e^{i\log i}$! Since $\log i = k2\pi i + \frac{\pi}2i$, we know that $i\log i = -k2\pi - \frac{\pi}2$. Negating $k$ for convenience, $i \log i = k2\pi - \frac{\pi}2$. Thus $$i^i = e^{i\log i} = e^{-\pi/2}e^{2k\pi}$$ And when taking the principle branch of log, we get $i^i = e^{-\pi/2}$.