# From Euler's identity: i^i=exp(-pi/2)= 0.2079 (rounded)

1. Jun 11, 2009

### SW VandeCarr

From Euler's identity: i^i=exp(-pi/2)= 0.2079 (rounded). I've always thought of this as an interesting result although I don't know of any particular significance or consequence of it. Is there any?

2. Jun 11, 2009

### disregardthat

Re: i^i

i^i does not have a specific value. $$i^i=e^{i^2(\frac{\pi}{2}+2\pi \cdot n) }=e^{-(\frac{\pi}{2}+2\pi \cdot n) }$$ for all integers n.

Last edited: Jun 12, 2009
3. Jun 11, 2009

### SW VandeCarr

Re: i^i

I'm using ^ as raising to a power except for 'exp' where exp(x) means e^x

exp(i pi)= -1

SQRT [exp(i pi) = SQRT (-1)

exp(i pi/2) = i

exp ((i^2) pi/2)) = i^i = exp (-pi/2) = 0.2079 (rounded)

Last edited: Jun 11, 2009
4. Jun 11, 2009

### arildno

Re: i^i

SW VandeCarr:

Jarle is also using it in that sense.

However, as he pointed out, the complex logarithm is a multi-valued mapping, in contrast to the real logarithm.

5. Jun 11, 2009

### SW VandeCarr

Re: i^i

Thanks, but the algebra is correct, is it not? Normally I don't see the +2pi.n term in texts.

Last edited: Jun 11, 2009
6. Jun 11, 2009

### HallsofIvy

Staff Emeritus
Re: i^i

Strange, I do!

7. Jun 11, 2009

### daviddoria

Re: i^i

SW VandeCarr - you can put "tex" tags around your equations and use latex syntax instead of defining all of your notation. It makes yours and everyone else's life easier :)

8. Jun 11, 2009

### SW VandeCarr

Re: i^i

Thanks daviddoria. I guess it's about time I started using latex if I'm going to be posting questions here.