## i^i = a real number?

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

$e^{xi}=cos(x)+sin(x)i$

If we substitute the value $\pi$ for x it turns out that

$e^{i\pi}=-1$

most of us already knew this wonderfull trick.

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

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

$e^{-\frac{\pi}{2}}=i^i$

Calculating the value of $e^{-\frac{\pi}{2}}$ it turns out that

$i^i=0,2078795763....$

Isn't that just the weirdest thing ever??
 PhysOrg.com mathematics news on PhysOrg.com >> Pendulum swings back on 350-year-old mathematical mystery>> Bayesian statistics theorem holds its own - but use with caution>> Math technique de-clutters cancer-cell data, revealing tumor evolution, treatment leads
 Blog Entries: 9 Recognitions: Homework Help Science Advisor It is a real number.And i wouldn't call anything in mathematics weird. Daniel.
 Recognitions: Homework Help 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 : $$i^i = e^{-\frac{1}{2}(2k + 1)\pi}$$, where k ranges over all integers. The principal value is for zero k.

Recognitions:
Homework Help

## i^i = a real number?

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

 Quote by 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 : $$i^i = e^{-\frac{1}{2}(2k + 1)\pi}$$, where k ranges over all integers. The principal value is for zero k.

I think you mean:

$$i^i = e^{-\frac{1}{2}(k + 1)\pi}$$

since it works for every k*(pi/2) with zero k beeing the principal value if i'm correct
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Nope.Compare the case k=0 and k=1,for simplicity. Daniel.
 I see...
 Recognitions: Homework Help Sorry, I was wrong. The general form should be $$i^i = e^{-\frac{1}{2}\pi(4k + 1)}$$.
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Nope,it can't be that one. Daniel.

Recognitions:
Homework Help
 Quote by dextercioby Nope,it can't be that one. Daniel.
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.

Blog Entries: 9
Recognitions:
Homework Help
 Quote by Curious3141 Sorry, I was wrong. The general form should be $$i^i = e^{-\frac{1}{2}\pi(4k + 1)}$$.
Alright

You said (see above)

$$i^{i}=e^{-\frac{1}{2}\pi(4k+1)}$$

I say

$$k=0$$ (1)

$$\mathbb{R}\ni i^{i}=e^{-\frac{\pi}{2}}$$ (2)

$$k=1$$ (3)

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

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

Daniel.
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Okay,now here's what u and Max Cohen wanted to write. $$i^{i}=e^{-\frac{\pi}{2}+2ki\pi} \ ,\ k\in\mathbb{Z}$$ Daniel.
 Recognitions: Homework Help 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.
 Blog Entries: 9 Recognitions: Homework Help Science Advisor 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... $$i^{i}=:a\in\mathbb{R}$$ You're telling me that $$a=e^{-\frac{\pi}{2}}=e^{-\frac{5\pi}{2}}=e^{-\frac{9\pi}{2}}=...$$ and that's profoundly incorrect. Daniel.

Recognitions:
Homework Help
 Quote by dextercioby $$i^{i}=:a\in\mathbb{R}$$ You're telling me that $$a=e^{-\frac{\pi}{2}}=e^{-\frac{5\pi}{2}}=e^{-\frac{9\pi}{2}}=...$$ and that's profoundly incorrect.
This isn't what he's saying. He's saying that you have many different choices for $$i^i:=e^{i\log{i}}$$. log is a multivalued function, $$\log{i}=\pi i/2+2\pi i k$$ for any integer k. The choice of k chooses the branch of log you are working with. So $$i^i=e^{i(\pi i/2+2\pi i k)}=e^{-\pi /2-2\pi k}$$ and the different values of k give different values of i^i, depending on which branch of log you're using. $$e^{-\frac{\pi}{2}},e^{-\frac{5\pi}{2}},e^{-\frac{9\pi}{2}},\ldots$$ are all valid answers for i^i, just as $$\pi i/2,5\pi i/2,9\pi i/2,\ldots$$ are all valid answers for log(i).
 Quote by shmoe This isn't what he's saying. He's saying that you have many different choices for $$i^i:=e^{i\log{i}}$$. log is a multivalued function, $$\log{i}=\pi i/2+2\pi i k$$ for any integer k. The choice of k chooses the branch of log you are working with. So $$i^i=e^{i(\pi i/2+2\pi i k)}=e^{-\pi /2-2\pi k}$$ and the different values of k give different values of i^i, depending on which branch of log you're using. $$e^{-\frac{\pi}{2}},e^{-\frac{5\pi}{2}},e^{-\frac{9\pi}{2}},\ldots$$ are all valid answers for i^i, just as $$\pi i/2,5\pi i/2,9\pi i/2,\ldots$$ are all valid answers for log(i).