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Question Number 103253 by Study last updated on 13/Jul/20
 (i)^(1/i) =?????
$$\:\sqrt[{{i}}]{{i}}=????? \\ $$
Answered by Dwaipayan Shikari last updated on 13/Jul/20
i^(1/i) =i^(−i)     ((1/i)=(i/i^2 )=−i)  e^(πi) =−1  e^((πi )/2) =i  e^(((πi)/2).(−i)) =i^(−i)   e^(π/2) =i^(−i) ⇒4.81077...  For regularaization  e^((π/2)+2πk) (k∈Z)
$${i}^{\frac{\mathrm{1}}{{i}}} ={i}^{−{i}} \:\:\:\:\left(\frac{\mathrm{1}}{{i}}=\frac{{i}}{{i}^{\mathrm{2}} }=−{i}\right) \\ $$$${e}^{\pi{i}} =−\mathrm{1} \\ $$$${e}^{\frac{\pi{i}\:}{\mathrm{2}}} ={i} \\ $$$${e}^{\frac{\pi{i}}{\mathrm{2}}.\left(−{i}\right)} ={i}^{−{i}} \\ $$$${e}^{\frac{\pi}{\mathrm{2}}} ={i}^{−{i}} \Rightarrow\mathrm{4}.\mathrm{81077}… \\ $$$$\mathrm{For}\:\mathrm{regularaization} \\ $$$$\mathrm{e}^{\frac{\pi}{\mathrm{2}}+\mathrm{2}\pi\mathrm{k}} \left(\mathrm{k}\in\mathbb{Z}\right) \\ $$
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
(e^((π/2)i) )^(−i) =(i)^(−i     ) ok
$$\left({e}^{\frac{\pi}{\mathrm{2}}{i}} \right)^{−{i}} =\left({i}\right)^{−{i}\:\:\:\:\:} {ok} \\ $$
Commented by Study last updated on 13/Jul/20
(1/i)=i^(−1)      ok
$$\frac{\mathrm{1}}{{i}}={i}^{−\mathrm{1}} \:\:\:\:\:{ok} \\ $$
Commented by Dwaipayan Shikari last updated on 13/Jul/20
(1/i)=(i/i^2 )=−i
$$\frac{\mathrm{1}}{{i}}=\frac{{i}}{{i}^{\mathrm{2}} }=−{i} \\ $$
Commented by Dwaipayan Shikari last updated on 13/Jul/20
it means   e^((π/2)i.(−i)) =(i)^(−i)
$${it}\:{means}\: \\ $$$${e}^{\frac{\pi}{\mathrm{2}}{i}.\left(−{i}\right)} =\left({i}\right)^{−{i}} \\ $$
Answered by abdomsup last updated on 13/Jul/20
=i^(1/i)  =i^(−i)  =(e^((iπ)/2) )^(−i)  =e^(π/2)
$$={i}^{\frac{\mathrm{1}}{{i}}} \:={i}^{−{i}} \:=\left({e}^{\frac{{i}\pi}{\mathrm{2}}} \right)^{−{i}} \:={e}^{\frac{\pi}{\mathrm{2}}} \\ $$
Answered by OlafThorendsen last updated on 13/Jul/20
z = (i)^(1/i)  = i^(1/i)   lnz = (1/i)lni = (1/i)lne^(i(π/2))  = (1/i)i(π/2)  lnz = (π/2) ⇒ z = e^(π/2)
$${z}\:=\:\sqrt[{{i}}]{{i}}\:=\:{i}^{\frac{\mathrm{1}}{{i}}} \\ $$$$\mathrm{ln}{z}\:=\:\frac{\mathrm{1}}{{i}}\mathrm{ln}{i}\:=\:\frac{\mathrm{1}}{{i}}\mathrm{ln}{e}^{{i}\frac{\pi}{\mathrm{2}}} \:=\:\frac{\mathrm{1}}{{i}}{i}\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{ln}{z}\:=\:\frac{\pi}{\mathrm{2}}\:\Rightarrow\:{z}\:=\:{e}^{\frac{\pi}{\mathrm{2}}} \\ $$

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