Really Complex

Using Euler's formula,
$e^{i\theta} = \cos\theta + i\sin\theta$ $\text{Put} \ \theta = \dfrac{\pi}{2}$ $e^{i\frac{\pi}{2}} = \cos\dfrac{\pi}{2} + i\sin\dfrac{\pi}{2}$ $e^{i\frac{\pi}{2}} = i$ $\text{Raising i-th power to both sides,}$ $e^{i^2\frac{\pi}{2}} = i^i = \sqrt{-1}^{\sqrt{-1}}$ $\sqrt{-1}^{\sqrt{-1}} = e^{-\dfrac{\pi}{2}} = 0.208 = \text{real number} \quad \quad ( \because i^2 = -1)$

Using the properties of logarithms and the identity $\ e^{i \pi} = -1$ : $\ Ln( i^i) = i Ln (i) = i Ln (-1)^{1/2} = \frac{i}{2} Ln (-1)$ $\ = \frac{i}{2} Ln (e^{i \pi }) = \frac{1}{2} i.i \pi Ln (e) = - \frac{1}{2} \pi$ $\therefore\ i^i = e^{ - \frac{1}{2} \pi }$

Euler was great...

Here is how I figured it out using simple algebra. Let $i^i = m$ for any m Then $i^{4i} = m^4$ $(i^4)^i = m^4$ $m^4 = 1$ Thus $m$ is definitely real.

It is equal to e raised to power -π/2

Let's put $i^{i} = a + bi$ . Then $i^{i}(a-bi) = a^2+b^2$ . Unless $b=0$ , either $i^{i}\times a$ or $i^{i}\times bi$ would be imaginary, which is impossible because $a^2+b^2$ is real. Therefore $b=0$ and $i^{i}$ is real.

it equals e^(-pi/2)

$i^i = e^{i\ln{i}} = e^{i\ln{e^{i\dfrac{\pi}{2}}}} = e^{i*i*\dfrac{\pi}{2}} = e^{-\dfrac{\pi}{2}} = 0.207879$

((-1)^(1/2))^((-1)^(1/2)) = i^i = 1