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 May 25th, 2017, 02:44 PM #1 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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?
 May 25th, 2017, 03:15 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,660 Thanks: 2635 Math Focus: Mainly analysis and algebra Well $i$ is detached from $\frac{a}{b}$, so it's not surprising that there's a discontinuity there is it?
 May 25th, 2017, 03:21 PM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 $i^i= e^{ln(i^i)}= e^{i ln(i)}$. since $i= e^{i\frac{\pi}{2}}$ so $ln(i)= i\frac{\pi}{2}$. Then $i ln(i)= -\frac{\pi}{2}$ and $i^i= e^{-\frac{\pi}{2}}$, a real number. (That is the "principal value". You can get others by adding $2\pi i$ to the $i\frac{\pi}{2}$.)
 May 27th, 2017, 05:43 AM #4 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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 10:02 AM.
 May 29th, 2017, 04:52 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,660 Thanks: 2635 Math Focus: Mainly analysis and algebra $$i^z = (e^{i\frac\pi2})^{a+bi} = e^{-b\frac\pi2}e^{ai\frac\pi2}$$ That's the whole complex plane.
 May 30th, 2017, 03:33 AM #6 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Except the origin of course. If you are right , we just found a simple equation for the entire complex plane sans the origin. Thanks from v8archie
September 29th, 2017, 05:32 AM   #7
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Joined: Jan 2015
From: Alabama

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That should have been
Quote:
 Originally Posted by Country Boy $\displaystyle i^i= e^{ln(i^i)}= e^{i ln(i)}$. 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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