i^i

**SW VandeCarr** · Jun11-09, 05:03 PM

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?

**Jarle** · Jun11-09, 05:16 PM

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

**SW VandeCarr** · Jun11-09, 05:35 PM

Originally Posted by Jarle

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

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)

**arildno** · Jun11-09, 05:52 PM

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.

**SW VandeCarr** · Jun11-09, 06:01 PM

Originally Posted by arildno

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.

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

**HallsofIvy** · Jun11-09, 06:31 PM

Strange, I do!

**daviddoria** · Jun11-09, 06:51 PM

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 :)

**SW VandeCarr** · Jun11-09, 07:06 PM

Originally Posted by daviddoria

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 :)

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