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May 25th, 2017, 03:44 PM   #1
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Where is $i^i$ ?

For $ \ \ \mathbb{a} \ \ , \ \ \mathbb{ b} \ \ € \ \ \mathbb{ Z} \ \ , \mathbb{ b} \ne 0 $

We know where $ \ \ i^ { \frac{a}{b} } \ \ $ is. It is on the unit circle centered at the origin in the complex plane.

It is not hard to imagine that for $ \ \ \mathbb{ a} \ \ , \mathbb{ b} \ \ € \ \ \mathbb{ R} \ \ , \ \ \mathbb{ b} \ne 0 \ \ , \ \ i^ { \frac{a}{b} } \ \ $ fills up all the points on the unit circle centered at the origin in the complex plane.

Now , $ i^i \ \ $ is not on this unit circle. It is on the real number line detached from the 2-dimensional unit circle. Seems like we lost a dimension.

What is going on here?
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May 25th, 2017, 04:15 PM   #2
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Well $i$ is detached from $\frac{a}{b}$, so it's not surprising that there's a discontinuity there is it?
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May 25th, 2017, 04:21 PM   #3
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. since so . Then and , a real number.

(That is the "principal value". You can get others by adding to the .)
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May 27th, 2017, 06:43 AM   #4
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Here is a follow-up question:

Consider

$f(x) = i^x$, $x \ \in\ \mathbb{C}$

The base is $i$ and the domain of the exponent $x$ is all complex numbers.

What is the range? Or to put it differently, we know for $\ x \ \in\ \mathbb{R} \ $ the output is on the unit circle centered at the origin in the complex plane. We also know the output can fall on the real axis, for example $\ f(i) = i^i$. Where else can the output wind up? Can the output be any point in the complex plane?

Can $ \ \ i^x = a + bi \ \ $ for all $ \ \ a$, $b \ \in \ \mathbb{R} $ except possibly when $a$, $b$ are both simultaneously $0$?

??


Last edited by skipjack; May 27th, 2017 at 11:02 AM.
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May 29th, 2017, 05:52 PM   #5
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$$i^z = (e^{i\frac\pi2})^{a+bi} = e^{-b\frac\pi2}e^{ai\frac\pi2}$$

That's the whole complex plane.
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May 30th, 2017, 04:33 AM   #6
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Except the origin of course. If you are right , we just found a simple equation for the entire complex plane sans the origin.
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September 29th, 2017, 06:32 AM   #7
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That should have been
Quote:
Originally Posted by Country Boy View Post
$\displaystyle i^i= e^{ln(i^i)}= e^{i ln(i)}$. since $\displaystyle i= e^{i\frac{\pi}{2}}$ so $\displaystyle ln(i)= i\frac{\pi}{2}$. Then $\displaystyle i ln(i)= -\frac{\pi}{2}$ and $\displaystyle i^i= e^{-\frac{\pi}{2}}$, a real number.

(That is the "principal value". You can get others by adding $\displaystyle 2\pi i$ to the $\displaystyle i\frac{\pi}{2}$.)
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