Question

Imaginary numbers, what is:
\[i^i\]
I know that i is defined as: \[i = \sqrt{-1}\]

So how does this work out?

Answer 1

e^(-π/2)

Answer 2

@ikram002p
Please explain :-).
How you got that I know that \(e = 2.7182818\) but how do you get that \(e^{\frac{-\pi}{2}} = i^i\), how do we prove it?

Answer 3

e^i theta=cos (theta)+i sin(theta) right ?

Answer 4

I haven't studied that far yet :o.

I guess I will have to learn trigonometry to learn how to do this.

Thank you though!

Answer 5

oh .. dnt need trigonometry,
its the complex num formula
from tylor expansions , the rest of it so easy :o

Answer 6

just keep up with me ok ?

Answer 7

I only know Algebra II...
Taylor's expansion is calculus.

Answer 8

i dnt wanna u to prove tylor or somthing , just
know this formula
that
e^i theta=cos (theta)+i sin(theta) ok ?
else is so easy
u can note in the second term there is "i sin(theta)"
so find theta s.t sin theta gives u 1 ,-1 ok ?

Answer 9

so if u note that
when
sin (theta)=1,-1
cos (theta) =0

Answer 10

so just take theta = π/2
e^i π/2=cos (π/2)+i sin(π/2)
e^i π/2=0+i(1)
e^i π/2=i

Answer 11

e^i π/2=i
raise to the power i

( e^i π/2)^i=i^i
e^(i*i π/2)=i^i.............(i*i=-1)
e^-π/2=i^i # done
:D

Answer 12

@ikram002p
I'm so sorry, to leave you half way :O my internet disconnected, sorry!

I understand what you mean, thanks :D

Answer 13

np :D