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i(iota)to the power i(iota).
z = ei@ (representation of complex number Z in ellulers form) z1 = ii (given) taking log both sides logz1 = ilogi .................1 now , logi = log(ei@) = log(eipi/2) = ipi/2 puting this value of logi in eq 1 logZ1 = i2(pi/2) = -pi/2 (i2 = -1) Z1 = e(-pi/2) , this is the required result
z = ei@ (representation of complex number Z in ellulers form)
z1 = ii (given)
taking log both sides
logz1 = ilogi .................1
= log(eipi/2)
= ipi/2
puting this value of logi in eq 1
logZ1 = i2(pi/2) = -pi/2 (i2 = -1)
Z1 = e(-pi/2) , this is the required result
Hehe log is only defined for non negative real no. :P :DAnd you solved it?
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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