what is the value of i exponent i where i = square root negative 1

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Manoj Gupta

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0.20788 exp(i*x) = cos(x) + i*sin(x) exp(i*Pi/2) = cos(Pi/2) + i*sin(Pi/2) = i square both the sides now. i^i = exp(i*i*Pi/2) = exp(-Pi/2) = 0.20788 Interesting fact is that i raised to the i-th power is actually a real number!
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i exponent i = (e^ipi/2)^i = e^(-pi/2)
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IIT Foundation and Expert Level in Computer Technology

i ka power odd number= -1 and i ka power even number=1 ex, i ka power 3=-1,and i ka power 4=1
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i=cos(pi/2)+i*sin(pi/2)=e^(i*pi/2) so,i=i^i=e^((i*pi/2)*i)=e^(-pi/2)
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A proven maths teacher with 99 percentile in NMAT Maths,98.5 percentile in CAT Maths and 90+ in 10th and 12th

As i(iota)^2=-1 so i exponent will be 1 as i=square root of -1
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i=e^(i.theta) and i=cos(pi/2) + i sin (pi/2) So i=e^(i. pi/2) i^i =e^((i. pi/2). i) = e^(i.i.pi/2) =e^(-pi/2)
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we know that e^(ix)= cos(x)+isin(x) where i=sqrt(-1) so e^i=cos(1)+isin(1) =0.5403+0.8415i so, i*e^i=i*(0.5403+0.8415i) =0.5403i+0.8415*i^2 =0.5403i-0.8415
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