i^i=???

[Hyperreality]

What is i^i?? And how do we find it?

[Hurkyl]

By the definitions!

For complex numbers, exponentiation is defined by:

z^w := \exp(w \log z)

For the principal value of the exponential, you use the principal value of the logarithm.

[dextercioby]

By the definitions!

For complex numbers, exponentiation is defined by:

z^w := \exp(w \log z)

For the principal value of the exponential, you use the principal value of the logarithm.

Quite remarkable,Hurkyl,that both i^{i} and i^{\frac{1}{i}} are real numbers.

Yap,sometimes mathematics offers surprises... :cool:

[Tide]

Alternate HINT:

i^i = \left(e^{i\pi /2}\right)^i

[dextercioby]

Alternate HINT:

i^i = \left(e^{i\pi /2}\right)^i

:rofl: That's not a "HINT",that's the SOLUTION!!Hurkyl gave a hint.Anyway,i hope it helps him... :rofl:

[futb0l]

Mathworld has a page on complex exponentation.. http://mathworld.wolfram.com/ComplexExponentiation.html

[tongos]

well, what stumps me is how to find like (i+5)^(i+5)?????

[Tide]

well, what stumps me is how to find like (i+5)^(i+5)?????

HINT: Follow the hints offered above! :smile:

[Tsss]

The fact that (a^b)^c=a^{bc} is not right with complex numbers.

As a matter of fact, let z a complex number,
e^z=e^{\frac{2i\pi z}{2i\pi}}=(e^{2i\pi})^{\frac{z}{2i\pi}}=1^{\frac{ z}{2i\pi}}=1
there is a problem.

[cronxeh]

or, if you are an engineer, just open matlab and do i^i will save you the valuable time :biggrin:

[Tide]

The fact that (a^b)^c=a^{bc} is not right with complex numbers.

As a matter of fact, let z a complex number,
e^z=e^{\frac{2i\pi z}{2i\pi}}=(e^{2i\pi})^{\frac{z}{2i\pi}}=1^{\frac{ z}{2i\pi}}=1
there is a problem.

That's slick but

1^{\frac{z}{2\pi i}}=1

only if 1 = e^{0i} (on the LHS) but you explicitly took 1 = e^{2\pi i} and used a different expression of 1 in your final step. 1^z will be 1 only if arg z = 0.

[learningphysics]

That's slick but

1^{\frac{z}{2\pi i}}=1

only if 1 = e^{0i} (on the LHS) but you explicitly took 1 = e^{2\pi i} and used a different expression of 1 in your final step. 1^z will be 1 only if arg z = 0.

So when manipulating complex numbers we can't simply make a substitution and say:
1^\frac{z}{2\pi i}=(e^{(0i)})^{\frac{z}{2\pi i}}=e^0=1 ??

Which step above is illegal when manipulating complex numbers?

Is there a webpage, which states these types of situations...

[Tide]

So when manipulating complex numbers we can't simply make a substitution and say:
1^\frac{z}{2\pi i}=(e^{(0i)})^{\frac{z}{2\pi i}}=e^0=1 ??

Which step above is illegal when manipulating complex numbers?

Is there a webpage, which states these types of situations...

That is a valid root of 1 and there is nothing wrong with what you did. Tsss's problem arose because he inconsistently expressed the same number in two different ways going from one step to the next.

\left(e^{0 i}\right)^{1/2} = e^0 = 1

and

\left(e^{2\pi i}\right)^{1/2} = e^{i\pi} = -1

are distinct roots of 1 but if I applied Tsss's method to the second I could find only one square root of 1. Obviously, squaring either 1 or -1 both give 1.

Here's an appropriate web page: http://home.earthlink.net/~djbach/paradox.html