Question

how can i proving that i^i so that i on exponent i will be a real number
- from i we know that i^2 = -1 or sqrt(-1)=i

- can you help me solve this and understand it too ?

Answer 1

\[ i = 0 + i = \cos \left( \pi \over 2\right) + i \cos \left( \pi \over 2\right) = e^{i {\pi \over 2}}\]

Answer 2

Use Euler Identities..

Answer 3

thank you very very much but in a mathbook have wrote that

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

- is this true ?

Answer 4

should be i*sin(pi/2), but he's correct ofc.

Answer 5

@jhonyy9 srs? exponentiate both sides by i

Answer 6

yep ..
\[ e^{i {\pi \over 2} \times i} = e^{- {\pi \over 2}}\]
but i'm not quite how to interpret these types of powers (complex) and irrational.

Answer 7

i^i = (e^ipi/2)^i
= e^-pi/2 = 0.207

Answer 8

One of the Euler Identity says:

\[\large e^{i \theta} = \cos(\theta) + i \sin(\theta)\]

Replace \(\theta\) by \(\frac{\pi}{2}\)

\[\large e^{i \cdot \frac{\pi}{2}} = i \implies (e^{i \cdot \frac{\pi}{2}})^i = i^i = Solve\]

Answer 9

if anyone is interested you can find it here
http://math.stackexchange.com/questions/189703/does-ii-and-i1-over-e-have-more-than-one-root-in-0-2-pi


P.S. also don't forget to summarize me

Answer 10

@experimentX so in your first answer there are right ,,0+i = cos(pi/2) +cos(pi/2)i" ???

Answer 11

No....

Answer 12

\[i = 0 + i = \cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2})\]

Answer 13

According to Euler Identity :

\[\cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2}) = e^{i \frac{\pi}{2}}\]

Answer 14

Sorry ... typo from my side.

Answer 15

ok !

thank you !

good luck

jhonyy9