Question

we know that sqrt(-1)=i
and there exist too : i^i = ? what is difficile to calculi

yes ? true /false ?

Answer 1

i to the power of i is mathematically valid, so True

Answer 2

\(e^{i\theta}=\cos(\theta)+i\sin(\theta)\\e^{i\frac{\pi}{2}}=\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})\\e^{i\frac{\pi}{2}}=0+i*1=i\\

(e^{i\frac{\pi}{2}})^i=i^i\\i^i=e^{i^2\frac{\pi}{2}}\\i^i=\dfrac{1}{e^{\frac{\pi}{2}}}\approx0.20788 \)

Answer 3

Very strange. One would think that \(i\) would need to be raised to a rational number (in reduced form) where the numerator is even, in order to get a negative number.

Answer 4

thank you @zzr0ck3r so where i get this exercise was given that

i^i = e^( - pi/2) so exactly in a previous form how you wrote it above too

Answer 5

It is pretty common derivation. I think it should follow any students introduction to Euler's formula.

Answer 6

Great result. Just as good as
e^iπ + 1 = 0

Answer 7

not even close :) One is the daddy and one is a weird-o child

Answer 8

imo