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Question Number 53688 by kwonjun1202 last updated on 25/Jan/19

$$i^i??$$$$Whisisit,,plzexplainit$$$$i=\sqrt{-1}$$

Commented by Abdo msup. last updated on 25/Jan/19

$$wehavei=e^{\frac{i\pi }{2}}\Rightarrow i^i=e^{-\frac{\pi }{2}}.$$

Answered by MJS last updated on 25/Jan/19

$$\mathrm{i}=1\times {\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}}$$$${\mathrm{i}}^{\mathrm{i}}=1^{\mathrm{i}}\times {\mathrm{e}}^{{\mathrm{i}}^2\frac{\pi }{2}}=1\times {\mathrm{e}}^{-\frac{\pi }{2}}={\mathrm{e}}^{-\frac{\pi }{2}}=\frac{1}{\sqrt{{\mathrm{e}}^{\pi }}}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jan/19

$$weknow{a}^x=e^{{xlog}_{e}a}$$$$now{i}^i=e^{{iLog}_{e}i}=e^{iLog(0+1\times i)}$$$$nowformula$$$${Log}_e(\alpha +i\beta )={log}_e(\alpha +i\beta )+i2\pi n$$$$=[{log}_e\sqrt{{\alpha }^2+{\beta }^2}+{itan}^{-1}\left(\frac{\beta }{\alpha }\right)]+i2\pi n$$$${Log}_e(\alpha +i\beta )$$$$=\frac{1}{2}{log}_e({\alpha }^2+{\beta }^2)+i\{2\pi n+{tan}^{-1}\left(\frac{\beta }{\alpha }\right)\}$$$$soasperproblem\alpha =0\beta =1$$$$sovalueof{Log}_e(0+i)$$$$=\frac{1}{2}{log}_e(0^2+1^2)+i\{2\pi n+{tan}^{-1}\left(\frac{1}{0}\right)\}$$$$=\frac{1}{2}\times 0+i\{2\pi n+\frac{\pi }{2}\}$$$$now$$$$i^i=e^{{iLog}_{e}i}$$$$=e^{i\left\{(2\pi n+\frac{\pi }{2})i\right\}}$$$$=e^{(2\pi n+\frac{\pi }{2})\times -1}$$$$=e^{-(2\pi n+\frac{\pi }{2})}$$$$=e^{-\frac{\pi }{2}}principlevalue$$$$$$

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