Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 103253 by Study last updated on 13/Jul/20

$$\sqrt[i]{i}=?????$$

Answered by Dwaipayan Shikari last updated on 13/Jul/20

$$i^{\frac{1}{i}}=i^{-i}(\frac{1}{i}=\frac{i}{i^2}=-i)$$$$e^{\pi i}=-1$$$$e^{\frac{\pi i}{2}}=i$$$$e^{\frac{\pi i}{2}.(-i)}=i^{-i}$$$$e^{\frac{\pi }{2}}=i^{-i}\Rightarrow 4.81077...$$$$\mathrm{For}\mathrm{regularaization}$$$${\mathrm{e}}^{\frac{\pi }{2}+2\pi \mathrm{k}}(\mathrm{k}\in \mathbb{Z})$$

Commented by Dwaipayan Shikari last updated on 13/Jul/20

approximately 4.810477380965351655473035666703833126390170874664534940020... (using the principal branch of the logarithm for complex exponentiation)

Commented by Study last updated on 13/Jul/20

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

Commented by Study last updated on 13/Jul/20

$$\frac{1}{i}=i^{-1}ok$$

Commented by Dwaipayan Shikari last updated on 13/Jul/20

$$\frac{1}{i}=\frac{i}{i^2}=-i$$

Commented by Dwaipayan Shikari last updated on 13/Jul/20

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

Answered by abdomsup last updated on 13/Jul/20

$$=i^{\frac{1}{i}}=i^{-i}={\left(e^{\frac{i\pi }{2}}\right)}^{-i}=e^{\frac{\pi }{2}}$$

Answered by OlafThorendsen last updated on 13/Jul/20

$$z=\sqrt[i]{i}=i^{\frac{1}{i}}$$$$\ln z=\frac{1}{i}\ln i=\frac{1}{i}\ln e^{i\frac{\pi }{2}}=\frac{1}{i}i\frac{\pi }{2}$$$$\ln z=\frac{\pi }{2}\Rightarrow z=e^{\frac{\pi }{2}}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com