Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 136355 by mohammad17 last updated on 21/Mar/21

Answered by Dwaipayan Shikari last updated on 21/Mar/21

$$i^i={\left(e^{\frac{\pi i}{2}}\right)}^i=e^{-\frac{\pi }{2}}=\mathrm{\Phi }$$$$i^{log\left(i\right)}={\left(e^{\frac{\pi }{2}i}\right)}^{\frac{\pi }{2}i}=e^{-\frac{{\pi }^2}{4}}=\kappa (log\left(i\right)=\frac{\pi }{2}i)$$$$\mathrm{\Phi }+\kappa =e^{-\frac{\pi }{2}}+e^{-\frac{{\pi }^2}{4}}$$

Commented by mohammad17 last updated on 21/Mar/21

$$thankyousir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com