i^i = a real number?

Max cohen

Take euler's formula for the identity of complex numbers:

[latex]e^{xi}=cos(x)+sin(x)i[/latex]

If we substitute the value [latex]\pi[/latex] for x it turns out that

[latex]e^{i\pi}=-1[/latex]

most of us already knew this wonderfull trick.

BUT if we substitute [latex]\frac{\pi}{2}[/latex] for x we get (because cos pi/2 = 0 and sin pi/2 = 1):
[latex]e^{\frac{\pi}{2}i}=i[/latex]

Now if you raise both sides of this identity to the power i, you obtain (since i^2 = -1):

[latex]e^{-\frac{\pi}{2}}=i^i[/latex]

Calculating the value of [latex]e^{-\frac{\pi}{2}}[/latex] it turns out that

[latex]i^i=0,2078795763....[/latex]

Isn't that just the weirdest thing ever??

dextercioby

It is a real number.And i wouldn't call anything in mathematics weird.

Daniel.

Curious3141

That's just the principal value. There are an infinite number of values for the expression i^i, all real.

The general form is given by :

[tex]i^i = e^{-\frac{1}{2}(2k + 1)\pi}[/tex], where k ranges over all integers. The principal value is for zero k.

mathwonk

are you sure of that formula? try k=1.

Max cohen

I think you mean:

[tex]i^i = e^{-\frac{1}{2}(k + 1)\pi}[/tex]

since it works for every k*(pi/2) with zero k beeing the principal value if i'm correct

dextercioby

Nope.Compare the case k=0 and k=1,for simplicity.

Daniel.

Max cohen

I see...

Curious3141

Sorry, I was wrong. The general form should be [tex]i^i = e^{-\frac{1}{2}\pi(4k + 1)}[/tex].

dextercioby

Nope,it can't be that one.

Daniel.

Curious3141

Why not ?

k = 0, obvious.

k = 1, exp (-5pi/2) = z.

z^(-i) = exp(5*i*pi/2) = exp(i*pi/2 + 2*i*pi) = i

and so forth.

dextercioby

Alright

You said (see above)

[tex] i^{i}=e^{-\frac{1}{2}\pi(4k+1)} [/tex]

I say

[tex] k=0 [/tex] (1)

[tex] \mathbb{R}\ni i^{i}=e^{-\frac{\pi}{2}} [/tex] (2)

[tex] k=1 [/tex] (3)

[tex]\mathbb{R}\ni i^{i}=e^{-\frac{5\pi}{2}}\neq e^{-\frac{\pi}{2}}\substack{(2)\\\displaystyle{=}} i^{i}\in\mathbb{R} [/tex] (4)

Do you see something fishy...? You're working with very real numbers...No more multivalued functions...

Daniel.

dextercioby

Okay,now here's what u and Max Cohen wanted to write.

[tex]i^{i}=e^{-\frac{\pi}{2}+2ki\pi} \ ,\ k\in\mathbb{Z}[/tex]

Daniel.

Curious3141

No, that's not what I wanted to write. I meant that i^i has an infinite number of real, distinct values as given by the general form. The form you wrote trivially gives only a single real value.

My first form was wrong because it generated values for (-i)^i as well. But this form is definitely right.

Complex exponentiation is multivalued, and yes, it can even give an infinite number of distinct real results.

dextercioby

Nope.Everything is in terms of real numbers.U use equality sign,and the reals have a funny way of behaving way when in the presence of the equality sign...

[tex] i^{i}=:a\in\mathbb{R} [/tex]

You're telling me that [tex] a=e^{-\frac{\pi}{2}}=e^{-\frac{5\pi}{2}}=e^{-\frac{9\pi}{2}}=... [/tex]

and that's profoundly incorrect.

Daniel.

shmoe

This isn't what he's saying. He's saying that you have many different choices for [tex]i^i:=e^{i\log{i}}[/tex]. log is a multivalued function, [tex]\log{i}=\pi i/2+2\pi i k[/tex] for any integer k. The choice of k chooses the branch of log you are working with. So [tex]i^i=e^{i(\pi i/2+2\pi i k)}=e^{-\pi /2-2\pi k}[/tex] and the different values of k give different values of i^i, depending on which branch of log you're using. [tex]e^{-\frac{\pi}{2}},e^{-\frac{5\pi}{2}},e^{-\frac{9\pi}{2}},\ldots[/tex] are all valid answers for i^i, just as [tex]\pi i/2,5\pi i/2,9\pi i/2,\ldots[/tex] are all valid answers for log(i).

Curious3141

^Exactly.

dextercioby

Alright.Point taken. I never thought of it this way before,so i have a reason to thank you.

Daniel.