Question

Just out of interest - i found a surprising result in complex number theory - evaluate i^i using the identity cos x + i sin x = e^ix where x is in radians.

Answer 1

you let x = pi/2
so e ^(i pi) = cos (pi/2) + i sin (pi/2)

Answer 2

that last line should be
e ^(i pi/2) = cos (pi/2) + i sin (pi/2)

Answer 3

so e ^(i pi) = 0 + i
e ^(i pi/2)^1 = i^i
i^i = e^(-pi/2) = approx 0.208 - a real number!

Answer 4

yes you get
\[e^{-\frac{\pi}{2}}\]

Answer 5

but of course that is the "principle branch" because there are infinitely many

Answer 6

since the representation is not unique
i.e. you can use
\[i=e^{\frac{5\pi}{2}}\]as well

Answer 7

or more generally
\[i=e^{\frac{(2n-1)\pi}{2}}\]

Answer 8

right : pi/2 + 2npi

Answer 9

but it is cool that all are real!

Answer 10

yea

Answer 11

and of course i made a typo both times by omitting the i!

Answer 12

oh well!!

Answer 13

also... you can use
\[i^i=e^{i\ln(i)}\] and by some miracle it turns out that
\[\ln(i)=\frac{\pi i}{2}\] giving the same result

Answer 14

well not really a miracle it says the same thing.

\[log(z)=\log(r)+i\log(\theta)\]

Answer 15

great stuff

Answer 16

wish i remembered more of it that is for sure. mostly gone from my brain. someone actually asked this as a question here not too long ago

Answer 17

right - math never fails to surprise.
and much is still unknown - more than is known, i suspect.

Answer 18

i had a professor once who believed that just like rational numbers have measure 0 in the reals so decidable problems have measure 0

Answer 19

interesting - well - back to the mundane - my wife wants to go shopping!! see u

Answer 20

yeah time for market for me as well.

Answer 21

i'm sorry i missed this, very interesting!