i(iota)to the power i(iota).

Question

vikas askiitian expert · Answer

z = e i@ (representation of complex number Z in ellulers form) z 1 = i i (given) taking log both sides logz 1 = ilogi .................1 now , logi = log(e i@ ) = log(e ipi/2 ) = ipi/2 puting this value of logi in eq 1 logZ 1 = i 2 (pi/2) = -pi/2 (i 2 = -1) Z 1 = e (-pi/2) , this is the required result

Tejas Paresh Lodaya · Answer

Hehe log is only defined for non negative real no. :P :D And you solved it?

Vikas soni · Answer

i= e^i(π÷2)Now,i^i= {e^(i(π÷2))}^i= e^(-(π÷2)) which is positive and real. So the upper provided solution is true

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i(iota)to the power i(iota).

i(iota)to the power i(iota).

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