Imaginary numbers

Algebra Level 3

For $i = \sqrt{-1}$ , find the principal value of the number $\large i^i$ .

e^{-\pi/2}

e^{\pi/2}

\frac{\sqrt{2}}{2}+i \frac{\sqrt{2}}{2}

i

2 solutions

M R
Feb 13, 2016

Polar coordinate system z (z=a+bi) is z=re^(iθ) a=rcos (θ) , b=rsin (θ) , r=sqrt (a^2+b^2) and θ=arccos (a/r) or θ=arcsin (b/r) or...

i=sqrt (-1) => i=e^(i*(pi/2)) [r=sqrt (0^2+1^2)=1 and θ=arccos (0/1)=+(pi/2)]

i^i =(e^(i (pi/2)))^i =e^((i^2) (pi/2))=e^(-pi/2)

:-)

Sergio Rodriguez
Jan 21, 2016

Since we know that i=[e^{i \frac{\pi}{2}}] By exponent rule we have [e^{i^{2} \frac{\pi}{2}}]=[e^{-\frac{\pi}{2}}]