Question

please solve iⁱ

Answer 1

Answer on Question #80637 – Math – Complex Analysis

Question

Solve iⁱ.

Solution

$$i^i = e^{\ln i^i} = e^{i\ln i}.$$

Since $\ln i = \ln |i| + i\left(\frac{\pi}{2} + 2\pi k\right)$ for $k \in \mathbb{Z}$,

$$i \ln i = i \left(\ln |i| + i\left(\frac{\pi}{2} + 2\pi k\right)\right) = i \left(\ln 1 + i\left(\frac{\pi}{2} + 2\pi k\right)\right) = i \left(0 + i\left(\frac{\pi}{2} + 2\pi k\right)\right) = i^2 \left(\frac{\pi}{2} + 2\pi k\right) = -\left(\frac{\pi}{2} + 2\pi k\right).$$

Therefore,

$$i^i = e^{\ln i^i} = e^{i\ln i} = e^{-\left(\frac{\pi}{2} + 2\pi k\right)} \text{ for } k \in \mathbb{Z}.$$

Answer: $i^i = e^{-\left(\frac{\pi}{2} + 2\pi k\right)}$ for $k \in \mathbb{Z}$.

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