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Question Number 151560 by peter frank last updated on 21/Aug/21

$$\mathrm{show}\mathrm{that}{\mathrm{i}}^{\mathrm{i}}\mathrm{is}\mathrm{always}\mathrm{real}$$

Answered by puissant last updated on 21/Aug/21

$$i^2=-1\Rightarrow i=\sqrt{-1}$$$$i^i={\left(\sqrt{-1}\right)}^i=e^{iln\left(\sqrt{-1}\right)}=e^{\frac{i}{2}{lni}^2}=e^{ilni}$$$$=e^{i\frac{\pi }{2}i}=e^{-\frac{\pi }{2}}\in \mathbb{R}..$$

Commented by peter frank last updated on 22/Aug/21

$$\mathrm{thank}\mathrm{you}$$

Answered by Olaf_Thorendsen last updated on 21/Aug/21

$$z=i^i$$$$\ln z=i\ln i=i\ln \left(e^{i\frac{\pi }{2}}\right)=i\left(i\frac{\pi }{2}\right)=-\frac{\pi }{2}$$$$z=e^{-\frac{\pi }{2}}\in \mathbb{R}.$$

Commented by peter frank last updated on 22/Aug/21

$$\mathrm{thank}\mathrm{you}$$

Answered by MJS_new last updated on 21/Aug/21

$${\mathrm{i}}^{\mathrm{i}}={\left({\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}}\right)}^{\mathrm{i}}={\mathrm{e}}^{{\mathrm{i}}^2\frac{\pi }{2}}={\mathrm{e}}^{-\frac{\pi }{2}}$$

Commented by peter frank last updated on 22/Aug/21

$$\mathrm{thank}\mathrm{you}$$

Answered by peter frank last updated on 22/Aug/21

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