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Question-136355




Question Number 136355 by mohammad17 last updated on 21/Mar/21
Answered by Dwaipayan Shikari last updated on 21/Mar/21
i^i =(e^((πi)/2) )^i =e^(−(π/2)) =Φ  i^(log(i)) =(e^((π/2)i) )^((π/2)i) =e^(−(π^2 /4)) =κ    (log(i)=(π/2)i)  Φ+κ=e^(−(π/2)) +e^(−(π^2 /4))
$${i}^{{i}} =\left({e}^{\frac{\pi{i}}{\mathrm{2}}} \right)^{{i}} ={e}^{−\frac{\pi}{\mathrm{2}}} =\Phi \\ $$$${i}^{{log}\left({i}\right)} =\left({e}^{\frac{\pi}{\mathrm{2}}{i}} \right)^{\frac{\pi}{\mathrm{2}}{i}} ={e}^{−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}} =\kappa\:\:\:\:\left({log}\left({i}\right)=\frac{\pi}{\mathrm{2}}{i}\right) \\ $$$$\Phi+\kappa={e}^{−\frac{\pi}{\mathrm{2}}} +{e}^{−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}} \\ $$
Commented by mohammad17 last updated on 21/Mar/21
thank you sir
$${thank}\:{you}\:{sir} \\ $$

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