Imaginary Unit to Power of Itself/Complete
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Theorem
- $i^i = \set {\exp \paren {\dfrac {4 k + 3} 2 \pi}: k \in \Z}$
where $i$ is the imaginary unit.
Proof
\(\ds i^i\) | \(=\) | \(\ds \exp \paren {i \ln \paren i}\) | Definition of Complex Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \exp \paren {i \paren {\ln 1 + i \paren {\dfrac \pi 2 + 2 k \pi} } }\) | Definition of Complex Natural Logarithm: for all $k \in \Z$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \exp \paren {i \paren {i \paren {\dfrac \pi 2 + 2 k \pi} } }\) | Logarithm of 1 is 0 | |||||||||||
\(\ds \) | \(=\) | \(\ds \exp \paren {i^2 \dfrac \pi 2 + 2 k \pi i^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \exp \paren {-\dfrac \pi 2 - 2 k \pi}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \exp \paren {\dfrac {3 \pi} 2 + 2 k \pi}\) | as $k$ ranges over all integers | |||||||||||
\(\ds \) | \(=\) | \(\ds \exp \paren {\dfrac {4 k + 3} 2 \pi}\) |
$\blacksquare$
Also presented as
This result can also be presented as:
- $i^i = \set {\exp \paren {- \dfrac \pi 2 \paren {4 k + 1} }: k \in \Z}$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 4$. Elementary Functions of a Complex Variable: Exercise $6 \ \text{(iv)}$
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {A}.21$: Euler ($\text {1707}$ – $\text {1783}$): Footnote $2$