+118651 i^i ???
\(e^{i\pi}=cos\pi+isin\pi\\ e^{i\pi}=-1\\ \sqrt{e^{i\pi}}=\sqrt{-1}\\ e^{\frac{i\pi}{2}}=i\\ (e^{\frac{i\pi}{2}})^i=i^i\\ (e^{\frac{i^2\pi}{2}})=i^i\\ (e^{\frac{-1\pi}{2}})=i^i\\ i^i=e^{\frac{-\pi}{2}}\\ \)
e^(-pi/2) = 0.2078795763507619 approximately
so,
\(i^i=0.2078795763507619 \;\; approximately\)
+26387 i^i
\(\begin{array}{|rcll|} \hline i^i &=& [(-1)^{\frac12}]^i \qquad | \qquad i = \sqrt{-1} = (-1)^{\frac12}\\ \mathbf{i^i} & \mathbf{=} & \mathbf{(-1)^{\frac i2}} \\ &=& e^{\frac i2 \cdot ln(-1) } \\\\ && \ln(z) = \ln(|z|) +i\cdot arg(z)\\\\ && \ln(-1+0\cdot i) = \ln(|-1+0\cdot i|) +i\cdot arg(-1+0\cdot i)\\ && \ln(-1) = \ln(|\sqrt{(-1)^2+0^2}|) +i\cdot \arctan\left(\frac{0}{-1} \right)\\ && \ln(-1) = \ln(|1|) +i\cdot \pi \qquad | \qquad \ln(|1|)=\ln(1) = 0 \\ && \ln(-1) = 0 +i\cdot \pi \\ && \ln(-1) = i\cdot \pi \\\\ i^i &=& e^{\frac i2 \cdot \ln(-1) } \\ &=& e^{\frac i2 \cdot i\cdot \pi } \\ &=& e^{\frac {i^2}2 \cdot \pi } \qquad | \qquad i^2 = -1\\ &=& e^{\frac {-1}2 \cdot \pi } \\ &=& e^{\frac {- \pi}2 } \\ \mathbf{i^i} &\mathbf{=}& \mathbf{ e^{-(\frac {\pi}2) } } \\ \mathbf{i^i} &\mathbf{=}& \mathbf{0.207879576350761908546955619834978770033877841631769608075\dots }\\ \hline \end{array} \)
