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

i^(i )  ??  Whis is it,, plz explain it  i=(√(−1))

$${i}^{{i}\:} \:?? \\ $$$${Whis}\:{is}\:{it},,\:{plz}\:{explain}\:{it} \\ $$$${i}=\sqrt{−\mathrm{1}} \\ $$

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

we have i =e^((iπ)/2)  ⇒i^i =e^(−(π/2)) .

$${we}\:{have}\:{i}\:={e}^{\frac{{i}\pi}{\mathrm{2}}} \:\Rightarrow{i}^{{i}} ={e}^{−\frac{\pi}{\mathrm{2}}} . \\ $$

Answered by MJS last updated on 25/Jan/19

i=1×e^(i(π/2))   i^i =1^i ×e^(i^2 (π/2)) =1×e^(−(π/2)) =e^(−(π/2)) =(1/(√e^π ))

$$\mathrm{i}=\mathrm{1}×\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \\ $$$$\mathrm{i}^{\mathrm{i}} =\mathrm{1}^{\mathrm{i}} ×\mathrm{e}^{\mathrm{i}^{\mathrm{2}} \frac{\pi}{\mathrm{2}}} =\mathrm{1}×\mathrm{e}^{−\frac{\pi}{\mathrm{2}}} =\mathrm{e}^{−\frac{\pi}{\mathrm{2}}} =\frac{\mathrm{1}}{\sqrt{\mathrm{e}^{\pi} }} \\ $$

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

we know a^x =e^(xlog_e a)   now  i^i =e^(iLog_e i) =e^(iLog(0+1×i))   now formula  Log_e (α+iβ)=log_e (α+iβ)+i2πn                             =[log_e (√(α^2 +β^2 )) +itan^(−1) ((β/α))]+i2πn  Log_e (α+iβ)  =(1/2)log_e (α^2 +β^2 )+i{2πn+tan^(−1) ((β/α))}  so as per problem α=0    β=1  so value of Log_e (0+i)  =(1/2)log_e (0^2 +1^2 )+i{2πn+tan^(−1) ((1/0))}  =(1/2)×0+i{2πn+(π/2)}  now   i^i =e^(iLog_e i)    =e^(i{(2πn+(π/2))i})   =e^((2πn+(π/2))×−1)   =e^(−(2πn+(π/2)))   =e^(−(π/2))   principle value

$${we}\:{know}\:{a}^{{x}} ={e}^{{xlog}_{{e}} {a}} \\ $$$${now}\:\:{i}^{{i}} ={e}^{{iLog}_{{e}} {i}} ={e}^{{iLog}\left(\mathrm{0}+\mathrm{1}×{i}\right)} \\ $$$${now}\:{formula} \\ $$$${Log}_{{e}} \left(\alpha+{i}\beta\right)={log}_{{e}} \left(\alpha+{i}\beta\right)+{i}\mathrm{2}\pi{n} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[{log}_{{e}} \sqrt{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }\:+{itan}^{−\mathrm{1}} \left(\frac{\beta}{\alpha}\right)\right]+{i}\mathrm{2}\pi{n} \\ $$$${Log}_{{e}} \left(\alpha+{i}\beta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{log}_{{e}} \left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)+{i}\left\{\mathrm{2}\pi{n}+{tan}^{−\mathrm{1}} \left(\frac{\beta}{\alpha}\right)\right\} \\ $$$${so}\:{as}\:{per}\:{problem}\:\alpha=\mathrm{0}\:\:\:\:\beta=\mathrm{1} \\ $$$${so}\:{value}\:{of}\:{Log}_{{e}} \left(\mathrm{0}+{i}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{log}_{{e}} \left(\mathrm{0}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} \right)+{i}\left\{\mathrm{2}\pi{n}+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{0}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{0}+{i}\left\{\mathrm{2}\pi{n}+\frac{\pi}{\mathrm{2}}\right\} \\ $$$${now}\: \\ $$$${i}^{{i}} ={e}^{{iLog}_{{e}} {i}} \\ $$$$\:={e}^{{i}\left\{\left(\mathrm{2}\pi{n}+\frac{\pi}{\mathrm{2}}\right){i}\right\}} \\ $$$$={e}^{\left(\mathrm{2}\pi{n}+\frac{\pi}{\mathrm{2}}\right)×−\mathrm{1}} \\ $$$$={e}^{−\left(\mathrm{2}\pi{n}+\frac{\pi}{\mathrm{2}}\right)} \\ $$$$={e}^{−\frac{\pi}{\mathrm{2}}} \:\:{principle}\:{value} \\ $$$$ \\ $$

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